DUKE 

UNIVERSITY 


LIBRARY 


vv  i\\S 


FOUR  DISCOVERERS  AND  FOUNDERS  OF  GEOMETRY 

(See  pages  4-90-498) 


PLANE  AND  SOLID 
GEOMETRY 


BY 

FLETCHER  DURELL,  Ph.D. 

HEAD  OP  THE  MATHEMATICAL  DEPARTMENT,  THE  LAWRENCEVILLE  SCHOOL 


NEW  YORK 

CHARLES  E.  MERRILL  CO, 


wm 


DURELL’S  MATHEMATICAL  SERIES 


r 


ARITHMETIC 

Two  Book  Series 
Elementary  Arithmetic 
Teachers’  Edition 
Advanced  Arithmetic 
Three  Book  Series 
Book  One 
Book  Two 
Book  Three 

ALGEBRA 

Two  Book  Course 
Book  One 
Book  Two 

Book  Two  with  Advanced  Work 
Introductory  Algebra 
School  Algebra 

GEOMETRY 

Plane  Geometry 
Solid  Geometry 
Plane  and  Solid  Geometry, 

TRIGONOMETRY 

Plane  Trigonometry  and  Tables 
Plane  and  Spherical  Trigonometry 
and  Tables.  • 

Plane  and  Spherical  Trigonometry 
with  Surveying  and  Tables 
Logarithmic  and  Trigonometric 
Tables 


Copyright.  1904,  by  Charles  E.  Merrill  Cft 


It  o o UV 


5/3.^ 


PREFACE 

One  of  the  main  purposes  in  writing  this  book  has 
been  to  try  to  present  the  subject  of  Geometry  so  that  the 
pupil  shall  understand  it  not  merely  as  a series  of  correct 
deductions,  but  shall  realize  the  value  and  meaning  of  its 
principles  as  well.  This  aspect  of  the  subject  has  been 
directly  presented  in  some  places,  and  it  is  hoped  that  it  per- 
vades and  shapes  the  presentation  in  all  places. 

Again,  teachers  of  Geometry  generally  agree  that  the 
most  difficult  part  of  their  work  lies  in  developing  in 
pupils  the  power  to  work  original  exercises.  The  second 
main  purpose  of  the  book  is  to  aid  in  the  solution  of  this 
difficulty  by  arranging  original  exercises  in  groups,  each 
of  the  earlier  groups  to  be  worked  by  a distinct  method. 
The  pupil  is  to  be  kept  working  at  each  of  these  groups 
till  he  masters  the  method  involved  in  it.  Later,  groups 
of  mixed  exercises  to  be  worked  by  various  methods  are 
given. 

In  the  current  exercises  at  the  bottom  of  the  page, 
only  such  exercises  are  used  as  can  readily  be  solved  in 
connection  with  the  daily  work.  All  difficult  originals  are 
included  in  the  groups  of  exercises  as  indicated  above. 

Similarly,  in  the  writer’s  opinion,  many  of  the  numeri- 

(iii) 


iv 


PREFACE 


cal  applications  of  geometry  call  for  special  methods  of 
solution,  and  the  thorough  treatment  of  such  exercises 
should  be  taken  up  separately  and  systematically.  [See 
pp.  304-318,  etc.]  In  the  daily  extempore  work  only  such 
numerical  problems  are  included  as  are  needed  to  make 
clear  and  definite  the  meanr'ng  and  value  of  the  geometric 
principles  considered. 

Every  attempt  has  been  made  to  create  and  cultivate 
the  heuristic  attitude  on  the  part  of  the  pupil.  This  has 
been  done  by  the  method  of  initiating  the  pupil  into 
original  work  described  above,  by  queries  in  the  course 
of  proofs,  and  also  at  the  bottom  of  different  pages,  and 
also  by  occasional  queries  in  the  course  of  the  text  where 
definitions  and  discussions  are  presented.  In  the  writer V 
opinion,  the  time  has  not  yet  come  for  the  purely  heuristic 
study  of  Geometry  in  most  schools,  but  it  is  all-important  to 
use  every  means  to  arouse  in  the  pupil  the  attitude  and 
energy  of  original  investigation  in  the  study  of  the  subject. 

In  other  respects,  the  aim  has  been  to  depart  as  little 
as  possible  from  the  methods  most  generally  used  at 
present  in  teaching  geometry. 

The  Practical  Applications  (Groups  88-91)  have  been 
drawn  from  many  sources,  but  the  author  wishes  to  ex- 
press his  especial  indebtedness  to  the  Committee  which  has 
collected  the  Real  Applied  Problems  published  from  time 
to  time  in  School  Science  and  Mathematics , and  of  which 
Professor  J.  F.  Millis  of  the  Francis  W.  Parker  School 
of  Chicago  is  the  chairman.  Page  360  is  due  almost  en- 
tirely to  Professor  William  Bets;  of  the  East  High  School 
of  Rochester,  N.  Y. 

FLETCHER  DURELL. 

Lawekncevulie.  N.  J. , Sept.  1,  1904. 


TO  THE  TEACHER 


1.  In  working  original  exercises,  one  of  the  chief  dif- 
, ficulties  of  pupils  lies  in  their  inability  to  construct  the 

figure  required  and  to  make  the  particular  enunciation 
from  it.  Many  pupils,  who  are  quite  uuable  to  do  this 
preliminary  work,  after  it  is  done  can  readily  discover  a 
proof  or  a solution.  In  many  exercises  in  this  book  the 
figure  is  drawn  and  the  particular  enunciation  made.  It 
is  left  to  the  discretion  of  the  teacher  to  determine  for 
what  other  exercises  it  is  best  to  do  this  for  pupils. 

2.  It  is  frequently  important  to  give  partial  aid  to  the 
pupil  by  eliciting  the  outline  of  a proof  by  questions  such 
as  the  following:  "On  this  figure  (or,  in  these  two  tri- 
angles) what  angles  are  equal,  and  why?”  "What  lines 
are  equal,  and  why?”  etc. 

3.  In  many  cases  it  is  also  helpful  to  mark  in  colored 
crayon  pairs  of  equal  lines,  or  of  equal  angles.  Thus,  in 
the  figure  on  p.  37  lines  AB  and  BE  may  be  drawn  with 
red  crayon,  AG  and  BE  with  blue,  and  the  angles  A and  D 
marked  by  sinaU  arcs  drawn  with  green  crayon.  If 


I colored  crayons  are  not  at  hand,  the  homologous  equa. 
parts  may  be  denoted  by  like  symbols  placed  on  them  . 
thus:  c F 


In  solving  theorems  concerning  proportional  lines,  it  is 
Occasionally  helpful  to  denote  the  lines  in  ~ proportion 


SYMBOLS  AND  ABBREVIATIONS 


+ plus,  or  increased  by. 

Adj.  . 

. adjacent. 

— minus,  or  diminished  by. 

Alt.  . 

. alternate. 

X multiplied  by. 

Art.  . 

. article. 

-4-  divided  by. 

Ax.  . . 

. axiom. 

— equals;  is  (or  are)  equal  to. 

Constr. 

. construction , 

= approaches  (as  a limit). 

Cor.  . 

. corollary. 

=o=  is  (or  are)  equivalent  to. 

Def.  . 

. definition. 

is  (or  are)  greater  than. 

Ex.  . . 

. exercise. 

<(  is  (or  are)  less  than. 

Ext.  . 

. exterior. 

.’.  therefcre. 

Fig.  . 

. figure. 

J.  perpendicular , perpendicular  to, 

Hyp.  . 

. hypothesis. 

or,  is  perpendicular  to. 

Ident. 

. identity. 

A perpendiculars. 

Int. 

. interior. 

||  parallel,  or,  is  parallel  to. 

Post.  . 

. postulate. 

||  s parallels. 

Prop.  . 

. proposition. 

Z , A angle,  angles 

Et.  . . 

. right. 

A,  A triangle,  triangles. 

Sug.  . 

. suggestion. 

CO , CO  parallelogram,  parallelograms. 

Sup.  . 

. supplemental 

G,  © circle,  circles. 

St.  . . 

. straight. 

Q.  E.  d.  quod  erat  demonstrandum;  that  is,  vrhieh  was  to  be 
proved. 

q.  E.  f.  quod  erat  faciendum;  that  is,  which  was  to  be  made. 

A few  other  abbreviations  and  symbols  will  be  introduced  and 
their  meaning  indicated  later  on. 


DEFINITIONS  AND  FIRST  PRINCIPLES 


INTRODUCTORY  ILLUSTRATIONS.  DEFINITIONS 

1.  Computation  of  an  area.  Practical  experience  has 
taught  men  that  certain  ways  of  dealing  with  objects  in 
the  world  about  us  are  more  advantageous  than  others; 
thus,  if  it  be  desired  to  find  the  num- 
ber of  square  yards  in  the  area  of  a 
floor,  we  do  not  mark  off  the  floor 
into  actual  square  yards  and  count 
the  number  of  square  yards  thus 
made,  but  pursue  the  much  easier 
course  of  measuring  two  lines,  the 
length  and  breadth  of  the  floor,  obtaining  7 yards,  say,  as 
the  length,  and  5 yards  as  the  width,  and  multiplying 
the  length  by  the  breadth.  The  area  is  thus  found  to  be 
35  square  yards. 

Let  the  pupil  determine  the  area  of  some  convenient  floor  in  each 
of  these  two  ways,  and  compare  the  labor  of  the  two  processes. 

2.  Computation  of  a volume.  Similarly,  for  example, 
in  order  to  determine  the  number  of  cubic  feet  which  a 
box  contains,  instead  of  filling  the  box  with  blocks  of 
wood,  each  of  the  size  of  a cubic  foot,  and 
counting  the  number  of  blocks,  we  pur- 
sue the  much  easier  course  of  measuring 
the  number  of  linear  feet  in  each  inside 
edge  of  the  box  and  multiplying  together 
the  three  dimensions  obtained;  thus,  if 

(9) 


1 

1 

1 

— 1 — 1 — 

! Li 

— 

1 

1 

— j— 1 
1 

1 

1 

1 

1 

— ,L“ 
1 

■""I-” 

1 

1 

■ j T 
! J 

1 

1 

■ » s 

i i ! 

10 


PLANE  GEOMETRY 


the  inside  dimensions  are  5,  4 and  3 feet,  the  volume  is 
5 X 4 X 3,  or  60  cubic  feet. 

Similarly,  the  direct  method  of  measuring  the  number  of  bushels 
of  wheat  in  a bin  is  to  fill  a bushel  measure  with  wheat  from  the 
bin,  time  after  time,  till  the  bin  is  exhausted,  and  count  the  number 
of  times  the  bushel  measure  is  used.  But  a much  less  laborious 
method  is  to  measure  the  three  dimensions  of  the  bin  in  inches 
and  divide  their  product  by  the  number  of  cubic  inches  in  a bushel. 

Let  the  pupil  in  like  manner  compare  the  labor  of  finding  the 
number  of  feet  of  lumber  in  a given  block  of  wood  by  actually  saw- 
ing the  block  up  into  lumber  feet,  with  the  labor  of  measuring  the 
dimensions  of  the  block  and  computing  the  number  of  lumber  feet 
by  taking  the  product  of  the  linear  dimensions  obtained. 

3.  Unknown  line  determined  from  known  lines.  The 

student  is  also  probably  familiar  with  the  fact  that,  by 
computations  based  on  the  relations  of  cer- 
tain lines  whose  lengths  are  known,  the 
lengths  of  other  unknown  lines  may  be 
determined  without  the  labor  of  measuring 
these  unknown  lines. 

Thus,  for  instance,  if  a ladder  5 yards  long  lean 
against  a wall  and  have  its  foot  3 yards  from  the 
wall,  the  height  of  the  top  of  the  ladder  from  the 
ground  may  be  determined  thus: 

(Height)2  = 52  — 32  = 25  — 9 = 16. 

.'.  height  = 4 yards. 

4.  Economies  in  representing  surfaces,  lines,  etc.  Other 

principles  of  advantage  of  an  even  more  general  character 
occur  in  dealing  with  geometric  objects.  Thus,  since  only 
one  straight  line  can  be  passed  through  two  given  points, 
the  two  points  may  be  taken  as  a highly  economized 
symbol  or  representative  of  the  line,  by  the  use  of  which 
much  labor  is  saved  in  dealing  with  lines,  and  new  results 
are  made  attainable. 


Geometric  magnitudes 


11 


Similarly,  since  only  one  flat,  or  plane  surface  can  be 
passed  through  three  given  points  (not  in  the  same  straight 
line),  these  three  points  may  be  taken  as  a symbol  or 
representative  of  the  flat  surface.  This  gives  the  advan- 
tage not  only  of  reducing  an  unlimited  surface  to  three 
points,  but  also  of  giving  for  the  plane  a symbol  made  up 
of  three  parts.  By  varying  one  of  these  parts  and  not  the 
others,  the  plane  may  be  varied  in  one  respect  and  not  in 
others;  also  planes  having  certain  properties  in  common 
may  be  grouped  together,  and  dealt  with  in  the  groups 
formed. 

5.  Geometry  as  a science.  Definition.  The  above  illus- 
trations serve  to  show  how  advantageous  it  often  is  to  deal 
with  geometric  magnitudes  by  certain  methods  rather  than 
others. 

In  the  study  of  Geometry  as  a science  we  proceed  to 
make  a systematic  examination  of  these  methods. 

Geometry  is  the  science  which  treats  of  the  properties  of 
continuous  magnitudes  and  of  space. 

GEOMETRIC  MAGNITUDES 

6.  Solids.  A physical  solid  is  a portion  of  matter,  as  a 
."block  of  wood,  an  iron  weight,  or  a piece  of  marble. 

The  portion  of  space  occupied  by  a physical  solid  may 
\be  considered  apart  from  the  physical  solid  itself,  hence 

A geometric  solid  is  the  portion  of  space  occupied  by  a 
physical  solid,  or  definitely  determined  in  any  way.  Hence, 
also,  a geometric  solid  is  a limited  portion  of  space. 

One  advantage  in  using  geometric  solids  lies  in  this.  If  we  dealt 
with  physical  solids  only,  as  blocks  of  wood,  marble,  iron,  etc.,  we 
should  need  to  determine  the  properties  of  each  kind  of  physical  solid, 


12 


PLANE  GEOMETRY 


separately;  but,  by  determining  the  properties  of  a geometric  solid, 
we  determine  once  for  all  the  properties  of  every  physical  solid,  no 
matter  what  its  material,  that  will  exactly  fill  the  space  occupied  by 
the  given  geometric  solid. 

Hereafter  in  this  book  the  term  "solid”  is  understood  tc 
mean  geometric  solid,  unless  it  be  otherwise  specified. 

7.  Other  geometric  magnitudes  defined  as  boundaries. 

A surface  is  the  boundary  of  a solid. 

A line  is  the  boundary  of  a surface. 

A point  is  the  boundary  of  a line. 

The  solid,  surface,  line,  and  point  are  the  fundamental 
geometric  magnitudes. 

8.  Geometric  magnitudes  defined  by  their  dimensions. 

A solid  has  three  dimensions;  viz.,  length,  breadth  and 
thickness. 

A surface  has  two  dimensions,  length  and  breadth. 

A line  has  one  dimension,  length. 

A point  has  no  dimension.  Hence  a point  is  that  which 
has  position,  but  no  magnitude. 

9.  Geometric  magnitudes  defined  as  generated  by  mution. 

A line  is  that  which  is  generated  by  the  motion  of  a point. 

A surface  is  that  which  is  generated  by  the  motion  of  a 

line  (not  moving  along  itself). 

A solid  is  that  which  is  generated  by  the  motion  of  a 
surface  (not  moving  along  itself). 

The  three  independent  motions  by  which  a solid  is  generated  illus 
trate  the  fact  that  a solid  has  three  dimensions. 

1Q„  Geometric  magnitudes  as  intersections.  The  inter- 
section of  two  surfaces  is  a line;  of  two  lines  is  a point. 

It  is  sometimes  more  advantageous  to  regard  geometric 
magnitudes  from  one  of  the  above  points  of  view,  some- 
times from  another. 


LINES 


13 


11.  A geometric  figure  is  any  combination  of  points, 
lines,  surfaces,  or  solids. 

12.  The  form  or  shape  of  a figure  is  determined  by  the 
relative  position  of  its  parts. 

13.  Similar  geometric  figures  are  those  which  have  the 
same  shape. 

Equivalent  figures  are  those  which  have  the  same  size. 

Equal  or  congruent  figures  are  those  which  have  the 
same  shape  and  size,  and  can,  therefore,  be  made  to 
coincide. 

14.  A point  is  represented  to  the  eye  by  a dot  and  is 
named  by  a letter  affixed  to  the  dot,  as  the  point  A,  • A. 


LINES 

15.  A straight  line  is  a line  such  that,  if  any  two  points 
in  it  be  fixed  and  the  line  l'otated,  every  point  in  the  line 
will  retain  its  original  position. 

A straight  line  is  also  sometimes  described  as  a line 
which  has  the  same  direction  throughout  its  whole  extent; 
or,  as  the  shortest  line  connecting  two  points. 

The  word  “line”  may  be  used  for  

“straight  line,”  if  no  ambiguity  results. 

16.  A curved  line  is  a line  no  portion 
of  which  is  straight.  The  word  "curve” 
is  often  used  for  "curved  line.” 

17.  A broken  line  is  a line  made  up  of  different  straight 
lines. 

18.  A rectilinear  figure  is  a figure  composed  only  of 
straight  lines;  a curvilinear  figure  is  a figure  composed 
only  of  curved  lines;  a mixtilinear  figure  is  a figure 
containing  both  straight  and  curved  lines. 


14 


PLANE  GEOMETKY 


19.  Kinds  of  straight  line.  A straight  line  may  be 
definite  or  indefinite  in  length. 

The  line  of  definite  length  is  sometimes  termed  a 
segment  or  sect. 

Other  kinds  of  straight  line  are  defined  in  Arts.  26  and  41. 

20.  Naming  a straight  line.  A straight  line  is  named 
by  naming  two  of  its  points,  as  the 
line  AB  (a  sect) ; or  the  line  CD  (in- 
definite in  length).  A segment  or  sect 
may  also  be  denoted  by  a single 
(small)  letter,  as  the  line  a. 

21.  The  circumference  of  a circle  is  a line  every  point 
of  which  is  equally  distant  from  a fixed  point  within, 
called  the  center. 


ANGLES 

22.  An  angle  is  the  amount  of  opening  between  two 
straight  lines  which  meet  at  a point. 

The  sides  of  an  angle  are  the  lines  whose  intersection 
forms  the  angle;  the  vertex  of  an  angle  is  the  point  in 
which  the  sides  intersect. 

23.  Naming  angles.  (1)  The  most  precise  way  of 
naming  an  angle  is  to  use  three  let- 
ters, one  for  a point  on  each  side  of 
the  angle  with  the  letter  at  the  vertex 
between  these  two,  as  the  angle  ABC. 

Since  the  size  of  an  angle  is  indepen- 
dent of  the  length  of  its  sides,  the  points 
named  on  its  sides  may  be  taken  at  any 
place  on  its  sides.  Thus,  the  angles  AOD, 

BOD , BOE,  AOF  are  all  the  same  angle. 


DBF 


ANGLES 


15 


(2)  In  case  there  is  but  one  angle  at  a given  vertex, 
the  letter  at  the  vertex  alone  is  sufficient  to  denote  the 
angle,  as  the  angle  0 in  the  last  figure. 

(3)  Sometimes  a letter  or  figure  placed  inside 
the  angle  and  near  the  vertex  is  a convenient 
symbol,  as  the  angle  a. 


24.  A straight  angle  is  an  angle  whose  sides  lie  in  the 
same  straight  line,  but  which  ex- 
tend in  opposite  directions  from  b . ^ 

the  vertex,  as  the  angle  BOA.  0 


25.  A right  angle  is  one  of  two  equal  angles  made  by 
one  straight  line  meeting  another  straight  line.  Thus,  if  the 
line  PQ  meets  line  AB  so  as  to  make 

angle  PQA  equal  to  angle  PQB,  each  of  P 

these  angles  is  a right  angle.  A right 
angle  is  also  half  of  a straight  angle. 

26.  A perpendicular  is  a line  that  B Q A 
makes  a right  angle  with  a given  line; 

thus  PQ  in  the  last  figui’e  is  perpendicular  to  BA.  The 
foot  of  a perpendicular  is  the  point  in  which  the  perpen- 
dicular meets  the  line  to  which  it  is  drawn,  thus  Q is  the 
foot  of  the  perpendicular  PQ. 


27.  An  acute  angle  is  an  angle 
.ess  than  a right  angle,  as  the 
angle  A 00. 


28.  An  obtuse  angle  is  an 
angle  greater  than  a right  angle 
but  less  than  a straight  angle,  as 
angle  A OD. 


D. 


■A 


O 


16 


- PLANE  GEOMETRY 


29.  A reflex  angle  is  an  angle 
greater  than  a straight  angle,  but  less 
than  two  straight  angles,  as  angle  A OF. 
In  this  book,  angles  larger  than  a straight 


angle  are  not  considered  unless  special  mention  is  made  of  them. 


30.  An  oblique  angle  is  an  angle  that  is  neither  a right 
nor  a straight  angle.  Hence,  oblique  angle  is  a general  term 
for  acute,  obtuse,  and  reflex  angles. 


31.  Adjacent  angles  are  angles 
which  have  a common  vertex  and  a 
common  side  between  them,  as  angles 
A OB  and  BOG. 


32.  Vertical  angles  are  angles 
which  have  a common  vertex  and 
the  sides  of  one  angle  the  prolonga- 
tions of  the  sides  of  the  other  angle, 
as  the  angles  AOC  and  BOD. 


33.  Complementary  angles  are  two  angles  which  to- 

gether  equal  a right  angle,  as  the  angles 
AOP  and  POQ. 

Hence,  the  complement  of  an  angle 
is  the  difference  between  that  angle 
and  one  right  angle. 

34.  Supplementary  angles  are  two 

angles  which  together  equal  two  right 
angles  (or  a straight  angle),  as  the  an- 
gles AOP  and  POP. 

Hence,  the  supplement  of  an  angle 
is  the  difference  between  that  angle  and  two  right 
angles. 


ANGLES 


17 


35.  Angles  as  formed  by  a rotating  straight  line.  If 

the  line  OB  start  in  the  position  OA  and  rotate  to  the 
position  OB,  it  is  said  to  generate  the 


D. 


E— 


F'' 


,B 


/V 


s 

A 


ZAOB. 

The  size  of  an  angle  may,  therefore, 
be  considered  as  the  amount  of  rotation 
of  a line  about  a point,  from  the  origi- 
nal position  of  the  line. 

If  the  rotation  is  continued  far 
enough,  a right  angle  (ZAOC)  is  formed;  afterward  an 
obtuse  angle  (ZAOB);  then  a straight  angle  (ZAOB), 
and  a reflex  angle  ( ZAOB) , etc. 

An  advantage  of  this  method  of  forming  or  conceiving 
angles  is  that  by  continuing  the  rotation  of  the  moving 
line  an  angle  of  indefinite  size  may  be  formed. 


36.  Units  of  angle.  A right  angle  is  a unit  of  angle 
useful  for  many  purposes.  Sometimes  a smaller  unit  of 
angle  is  needed. 

A degree  is  one -ninetieth  of  a right  angle;  a minute  is 
one-sixtieth  of  a degree;  a second  is  one-sixtieth  of  a 
minute. 

Besides  these,  other  units  of  angle  are  used  for  certain  purposes. 


SURFACES.  DIVISIONS  OF  GEOMETRY. 

PARALLEL  LINES 

37,  A plane  is  a surface  such  that,  if  any  two  points  in 
the  surface  be  joined  by  a straight  line,  the  straight  line 
lies  wholly  in  the  surface. 

Hence,  a plane  figure  is  a figure  such  that  all  its  points 
lie  in  the  same  plane. 

38.  A curved  surface  is  a surface  no  part  of  which  is 
plane. 

B 


18 


PLANE  GEOMETRY 


39.  Plane  Geometry  is  that  branch  of  Geometry  which 
treats  of  plane  figures. 

40.  Solid  Geometry  is  that  branch  of  Geometry  which 
treats  of  figures  all  points  of  which  are  not  in  the  same 
plane. 

41.  Parallel  lines  are  straight  lines  in  the  same  plane 
which  do  not  meet,  however  far  they  be  produced. 

EXERCISES.  CROUP  I 

Ex.  1.  Draw  a straight  line.  A curved  line.  A broken  line. 

Ex.  2.  Which  of  the  capital  letters  of  the  alphabet  are  straight, 
which  curved,  which  broken,  which  curved  and  straight  lines 
combined  ? 

Ex.  3.  Draw  an  acute  angle.  An  obtuse  angle.  A reflex  angle. 

Ex.  4.  Draw  two  adjacent  angles.  Two  vertical  angles. 

Ex.  5.  What  is  the  complement  of  an  angle  of  43°  ? What  is  its 
supplement  ? 

Ex.  6.  What  is  the  complement  of  57°  19'  ? of  62°  23'  43"  ? What 
is  the  supplement  of  each  of  these  ? 

Ex.  7.  At  two  o'clock,  what  is  the  angle  made  by  the  hour  and 
minute  hands  of  a clock  ? at  three  o’clock  ? at  five  o’clock  ? 

Ex.  8.  At  1:30  o’clock,  what  angle  do  the  hands  of  a clock  make  ? 
at  2:15  ? at  8:45  ? 

Ex.  9.  What  kind  of  an  angle  is  the  supplement  of  an  obtuse 
angle?  of  an  acute  angle?  of  a right  angle? 

Ex.  10.  What  kind  of  a surface  is  the  floor  of  a room?  the  surface 
of  a baseball?  the  surface  of  an  egg?  the  surface  of  a hemisphere? 

Ex.  11.  If  the  surface  ABCD  move 
to  the  right,  what  solid  is  generated  by 
it?  What  surfaces  are  generated  by  its 
bounding  lines?  What  lines  by  the  ver- 
tices of  its  angles? 


GEOMETRIC  MAGNITUDES 


19 


Ex.  12.  Draw  two  supplementary  adjacent  angles.  Also  two  sup- 
plementary angles  that  are  not  adjacent.  Also  two  adjacent  angles 
that  are  not  supplementary. 

Ex.  13.  The  sum  of  a right  angle  and  an  acute  angle  is  what  kind 
of  an  angle?  Their  difference  is  what  kind  of  an  angle? 

Ex.  14.  The  sum  of  an  obtuse  angle  and  a right  angle  is  what 
kind  of  an  angle?  Their  difference  is  what  kind? 

Ex.  15.  If  three  straight  lines  meet  (but  do  not  intersect)  at  a 
point  in  a plane,  how  many  angles  have  this  point  as  their  common 
vertex?  Draw  a figure  illustrating  this  and  name  the  angles. 

Ex.  16.  How  many  angles  are  formed  if  four  lines  meet  (but  do 
not  intersect)  at  a point  in  a plane? 

Ex.  17.  How  many  degrees  are  in  an  angle  which  equals  twice  its 
complement? 

Ex.  18.  How  many  degrees  in  an  angle  which  equals  one-third 
its  supplement? 

Ex.  19.  What  kind  of  an  angle  is  greater  than  its  supplement? 
What  kind  is  less? 


PRIMARY  RELATIONS  OF  GEOMETRIC  MAGNITUDES 

42.  Certain  primary  relations  of  geometric  objects  have 
already  been  given  in  the  definitions  used  for  geometric 
objects.  We  now  proceed  to  investigate  the  relations  of 
geometric  magnitudes  more  generally  and  systematically. 

43.  An  axiom  is  a truth  accepted  as  requiring  no 
demonstration. 

44.  Two  kinds  of  axioms  are  used  in  geometry: 

1.  General  axioms,  or  axioms  which  apply  to  other 
kinds  of  quantity  as  well  as  to  geometric  magnitudes;  for 
instance,  to  numbers,  forces,  masses,  etc. 


2U  ' PLANE  GEOMETRY 

2.  Geometric  axioms,  or  axioms  which  apply  to  geo- 
metric magnitudes  alone. 

45.  The  general  axioms  may  be  stated  as  follows: 
GENERAL  AXIOMS 

1.  Things  which  are  equal  to  the  same  thing,  or  to  equal 
things,  are  equal  to  each  other. 

2.  If  equals  be  added  to  equals,  the  sums  are  equal. 

3.  If  equals  be  subtracted  from  equals,  the  remainders 
are  equal. 

4.  Doubles  of  equals  are  equal;  or,  in  general,  if  equals 
be  multiplied  by  equals  the  products  are  equal. 

5 Halves  of  equals  are  equal;  or,  in  general,  if  equals 
be  divided  by  equals  the  quotients  are  equal. 

6.  The  whole  is  equal  to  the  sum  of  its  parts. 

7.  The  whole  is  greater  than  any  of  its  parts. 

8.  A quantity  may  be  substituted  for  its  equal  in  any 
process. 

9.  If  equals  be  added  to,  or  subtracted  from,  unequals,  the 
results  are  unequal  in  the  same  order;  if  unequals  be  added 
to  unequals  in  the  same  order,  the  results  are  unequal  in 
that  order. 

10.  Doubles,  or  halves,  of  unequals  are  unequal  in  the 
same  order. 

11.  If  unequals  be  subtracted  from  equals,  the  remainders 
are  unequal  in  the  reverse  order. 

12.  If,  of  three  quantities,  the  first  is  greater  than  the 
second,  and  the  second  is  greater  than  the  third r then  the 
first  is  greater  than  the  thirds 


GEOMETRIC  AXIOMS 


21 


46.  The  value  of  the  general  axioms.  The  axioms  given 
above  seem  so  obvious  that  the  student  at  first  is  not  likely 
to  realize  their  value.  This  value  may  be  illustrated  as 
follows: 

If  the  distance  from  Washington  to  Philadelphia  be  known,  and 
also  the  distance  from  Philadelphia  to  New  York,  the  distance  from 
Washington  to  New  York  may  be  obtained  by  adding  together  the  two 
(distances  named;  for,  by  axiom  6,  the  whole  is  equal  to  the  sum  of 
its  parts.  Thus  the  labor  of  actually  measuring  the  distance  from 
Washington  to  New  York  is  saved. 

Again,  if  the  height  of  a schoolboy  of  a given  age  in  Paris  be 
measured,  and  the  height  of  a like  schoolboy  in  New  York  be  meas- 
ured, and  the  result  of  the  measurement  in  each  case  is  the  same,  we 
know  that  the  boys  are  of  the  same  height,  without  the  labor  and  cost 
of  bringing  the  boys  together  and  comparing  their  heights  directly  ; for, 
by  axiom  1,  things  which  are  equal  to  the  same  thing  are  equal  to 
each  other. 

Thus  the  general  axioms  are  to  be  looked  at  not  merely 
as  fundamental  equivalences,  but  also  as  fundamental 
economies.  For  many  purposes  the  latter  point  of  view  is 
more  important  than  the  former. 

47.  The  geometric  axioms  may  be  stated  as  follows : 

GEOMETRIC  AXIOMS 

1.  Through  two  given  points  only  one  straight  line  can 
he  passed. 

2.  A geometric  figure  may  he  freely  moved  in  space  with 
out  any  change  in  form  or  size. 

This  axiom  is  equivalent  to  regarding  space  as  uniform,  or  lioma  - 
loidal ; that  is,  as  having  the  same  properties  in  all  its  parts.  It  has 
already  been  assumed  in  some  of  the  definitions  given.  See  Arts.  15 
and  35. 

3.  Through  a given  point  one  straight  line  and  only  one 
can  he  drawn  parallel  to  another  given  straight  line. 

By  Art.  13,  geometric  figures  tvhich  coincide  are  equal. 


22 


PLANE  GEOMETEY 

48.  Utility  or  uses  of  the  geometric  axioms.  By  means 
of  the  first  geometric  axiom  we  are  able  to  shrink  or  con- 
dense any  straight  line  into  two  points.  Later  this  advan- 
tage gives  rise  to  many  other  advantages.  By  the  second 
geometric  axiom,  the  knowledge  which  we  have  of  one  geo- 
metric object  may  be  transferred  to  another  like  object,  how- 
ever widely  separated  in  space.  The  utility  of  the  third 
geometric  axiom  can  be  made  more  evident  when  we  come  to 
use  it  in  proving  new  geometric  truths. 

49.  A postulate  in  geometry  is  a construction  of  a geo- 
metric figure  admitted  as  possible. 

50.  The  postulates  of  geometry  may  be  stated  as  follows: 

1.  Through  any  two  points  a straight  line  may  be  drawn. 

2.  A straight  line  may  be  extended  indefinitely , or  it  may 
be  limited  at  any  point. 

3.  A circumference  may  be  described  about  any  given 
point  as  center , and  with  any  given  radius. 

These  postulates  limit  the  pupil  to  the  use  of  the  straight-edged 
ruler  and  the  compasses  in  constructing  figures  in  geometry.  Odc  of 
the  objects  of  the  study  of  geometry  is  to  discover  what  geometric 
figures  can  be  constructed  by  a combination  of  the  elementary  con- 
structions allowed  in  the  postulates;  that  is,  by  the  use  of  the  two 
simplest  drawing  instruments. 

5 1 . Logical  postulates.  Besides  the  postulates  which  are 
used  in  the  actual  construction  of  figures,  there  are  certain  , 
other  postulates  which  are  used  only  in  the  processes  of 
reasoning.  Thus,  for  purposes  of  reasoning,  a given 
angle  may  be  regarded  as  divided  into  any  convenient  num- 
ber of  equal  parts.  Whether  it  is  possible  actually  thus  to 
divide  this  angle  on  paper  by  use  of  the  ruler  and  com- 
passes, is  another  question. 


DEMONSTRATION  OF  PROPERTIES 


23 


DEMONSTRATION  OF  GEOMETRIC  RELATIONS 

52.  A geometric  proof,  or  demonstration,  is  a course  of 
reasoning  by  which  a relation  between  geometric  objects  is 
established. 

53.  A geometric  theorem  is  a statement  of  a truth  con- 
cerning geometric  objects  which  requires  demonstration. 

Ex.  The  sum  of  the  angles  of  a triangle  equals  two  right  angles. 

54.  A geometric  problem  is  a statement  of  the  construc- 
tion of  a geometric  figure  which  is  required  to  be  made. 

Ex.  On  a given  line  to  construct  a triangle  containing  three  equal 
angles. 

55.  A proposition  is  a general  term  for  either  a theorem 
or  a problem. 

Thus,  propositions  are  subdivided  into  two  classes: 
1,  Theorems;  2,  Problems. 

56.  Immediate  inference  is  of  two  kinds: 

1.  Changing  the  point  of  view  in  a given  statement 
Thus  the  statement,  "two  straight  lines  drawn  through 
two  given  points  must  coincide,”  may  be  changed  to  "two 
straight  lines  cannot  inclose  a space.” 

2.  Reasoning  which  involves  but  a single  step. 

Ex.  "All  straight  angles  are  equal;” 

.’.  "All  right  angles  are  equal.”  (Ax.  5.) 

57.  A corollary  is  a truth  obtained  by  immediate  infer- 
ence from  another  truth  just  stated  or  proved. 

58.  A scholium  is  a remark  made  upon  some  particular 
feature  of  a proposition,  or  upon  two  or  more  propositions 
which  are  compared. 


24 


PLANE  GEOMETRY 


59.  Hypothesis  and  conclusion.  A proposition  consists 
of  two  parts: 

1.  The  hypothesis,  or  that  which  is  known  or  granted. 

2.  The  conclusion,  or  that  which  is  to  be  proved  or  con- 
structed. 

Thus,  in  the  proposition,  "if  two  straight  lines  are  per- 
pendicular to  the  same  line,  they  are  parallel,”  the 
'hypothesis  is,  that  two  given  lines  are  perpendicular  to 
another  given  line;  the  conclusion  is,  that  the  two  given 
lines  are  parallel. 

60.  The  converse  of  a proposition  is  another  proposi- 
tion formed  by  interchanging  the  hypothesis  and  the 
conclusion  of  the  original  proposition. 

Thus,  theorem,  "every  point  in  the  perpendicular  bisector  of  a line 
is  equidistant  from  the  extremities  of  the  line;” 

Converse,  " every  point  equidistant  from  the  extremities  of  a line 
lies  in  the  perpendicular  bisector  of  the  line.” 

Or,  in  general,  theorem,  "if  A is  B,  then  X is  Y;” 
converse,  "if  X is  Y,  then  A is  B;” 

Frequently  a converse  is  formed  by  interchanging  part  only  of  an 
hypothesis  with  part  or  all  of  the  conclusion,  or  vice  versa. 

Thus,  theorem,  "if  A is  B and  C is  D,  then  X is  Y;” 
converse,  "if  A is  B and  X is  Y,  then  C is  D.” 

The  converse  of  a theorem  is  not  necessarilj-  true.  Thus, 
it  is  true  that  all  right  angles  are  equal,  but  it  is  not  true 
that  all  equal  angles  are  right  angles. 

61.  The  opposite  of  a theorem  is  a theorem  formed  by 
making  both  the  hypothesis  and  the  conclusion  of  the 
'original  theorem  negative. 

Ex.  theorem,  "if  A is  B,  then  X is  Y;n 

opposite,  "if  A is  not  B,  then  X is  not  Y” 


DEMONSTRATION  OF  PROPERTIES  25 

If  a direct  theorem  and  its  opposite  are  both  true,  the  converse  is 
known  to  be  true  without  proof. 

Also,  if  a theorem  and  its  converse  are  both  true,  the  opposite  is 
known  to  be  true  without  proof. 

Thus  certain  economies  arise  in  the  demonstration  of  theorems. 

62.  Methods  of  geometric  proof.  Several  principal 
methods  of  proving  theorems  are  used  in  geometry. 

1.  Direct  demonstration. 

2.  Proof  by  superposition,  in  which  two  figures  are  proved 
equal  by  making  one  of  them  coincide  with  the  other. 

3.  Indirect  demonstration,  which  consists  essentially  in 
showing  that  a given  statement  is  time  by  showing  that  its 
negative  cannot  be  true. 

Other  special  methods  of  proof  will  be  pointed  out  as 
they  occur  in  the  course  of  the  work. 

63.  Form  of  a proof.  The  statement  of  a theorem  and 
its  proof  consist  of  certain  distinct  parts  which  it  is  impor- 
tant to  keep  clearly  in  mind.  These  parts  are: 

1.  The  general  enunciation,  which  is  the  statement  of 
the  theorem  in  general  terms. 

2.  The  particular  enunciation,  or  statement  of  the 
theorem  as  applied  to  a particular  figure  used  to  aid  the 
mind  in  carrying  forward  the  proof. 

3.  The  construction  of  supplementary  parts  of  the  figure 
(not  necessary  in  all  proofs). 

4.  The  proof.  This  must  include  a reason  for  every 
statement. 

5.  The  conclusion. 

The  letters  Q.  E.  d.,  standing  for  "quod  erat  demon- 
strandum,” and  meaning  "which  was  to  be  proved,”  aro 
usually  annexed  at  the  end  of  a completed  demonstration. 


26 


PLANE  GEOMETRY 


EXERCISES.  CROUP  2 


Ex.  1.  In  the  figure  meas- 
ure AB;  then  measure  BC.  A — -Q 

Now  find  AC  without  meas- 
uring it.  What  axiom  have  you  used  1 

Ex.  2.  If  Z AOB= 60°,  Z BOC=  80°  and 
Z COD= 130°;  find  without  measuring  them  Z 
AOC  and  BOD  (reflex).  What  axiom  have  you 
used  ? 

Ex.  3.  Prove  by  repeated  use  of  the  first 
part  of  Axiom  1 that  magnitudes  equal  to  equal 
magnitudes  are  equal  to  each  other;  (thus, 
given  A=x,  B=y,  and  x=y.  Prove  A=B). 


Ex.  4.  Give  a numerical  illustration  of  Axiom  11. 

Ex.  5.  Show  by  the  axioms  that  a part  is  equal  to  a whole  di- 
minished by  the  remaining  part. 


Ex.  6.  Show  that  Axiom  1 is  a special  case  of  Axiom  8. 

Ex.  7.  If  a — x + y and  x = y,  show  by  use  of  the  axioms  that 
x r ia. 

Ex.  8.  Draw  a line  and  produce  it  so  that  the  produced  part  shall 
equal  another  given  line. 


Ex.  9.  By  use  of  the  compasses,  mark  off  on  a given  line  a part 
twice  as  long  as  another  given  line. 

Ex.  10.  On  a given  line  mark  off,  by  fewest  uses  of  the  compasses, 
a part  four  times  as  long  as  another  given  line. 

Ex.  1 1 . Draw  three  straight  lines  and  denote  them  by  l,  m. 
and  n.  Then  draw  a line  l + m — n,  and  also  a line  l — 2vi-j-3n. 


PROPERTIES  OF  LINES  INFERRED  IMMEDIATELY 

64.  If  two  straight  lines  have  two  points  in  common., 
the  lines  coincide  throughout  their  ivhole  extent  (Art.  47, 
Geom.  As.  1). 

Hence,  two  straight  lines  can  intersect  in  but  one  point. 

65.  If  two  straight  lines  coincide  in  party  they  coincide 
throughout. 


PROPERTIES  OF  ANGLES 


27 


86.  Only  one  straight  line  can  he  drawn  connecting  two 
given  points. 

, 67.  Two  straight  lines  cannot  enclose  a surface. 

68.  A given  straight  line  {sect)  can  he  divided  into  two 
vqual  parts  at  but  one  point. 

For  (by  Ax.  5)  halves  of  equals  (or  of  the  same  thing) 
are  equal. 

PROPERTIES  OF  ANGLES  INFERRED  IMMEDIATELY 

69.  All  straight  angles  are  equal. 

70.  A straight  angle  can  he  divided  into  tivo  equal 
angles  by  hut  one  line  at  a given  point 

in  the  given  straight  line. 

For  (Ax.  5)  halves  of  the  same  mag- 
nitude are  equal.  

71.  Hence,  at  a given  point  in  a straight  line  hut  one 
perpendicular  can  be  erected  to  the  line. 

72.  All  right  angles  are  equal. 

For  all  straight  angles  are  equal  (Art.  69)  and  halves 
of  equals  are  equal  (Ax.  5). 

73.  The  sum  of  the  two  adjacent  angles 
formed  by  one  straight  line  meeting  an- 
other straight  line  equals  tivo  right  angles. 

For  the  angles  formed  are  supplementary  adjacent 
.angles  (Arts.  31,  34). 

74.  If  tivo  adjacent  angles  are  together  equal  to  a 
straight  angle  {or  two  right  angles),  their  exterior  sides 
form  one  and  the  same  straight  line. 

For  their  exterior  sides  form  a straight  angle,  and 
hence  must  lie  in  a straight  line  (Art.  24). 


Fig'.  1 


28 


PLANE  GEOMETRY 


75.  The  complements  of  two  equal  angles  are  equal 
(Art.  72  and  Ax.  3);  the  supplements 
of  tivo  equal  angles  are  equal  (Art.  69, 

Ax.  3). 

76.  The  sum  of  all  the  angles  about  a 
point  equals  four  right  angles. 

Thus,  Za  + Z&  + Zc  + Zd  + Ze  — 

4 rt.  Z . 


77.  The  sum  of  all  the  angles  about 
a point  on  the  same  side  of  a straight 
line  passing  through  the  point  equals 
two  right  angles. 


Fig.  3 


Thus,  Z.p  + Z q + Z r = 2 rt.  Z. 


EXERCISES.  CROUP  3 


Ex.  1.  How  many  different  straight  lines  are  determined  by  three 
points  not  in  the  same  straight  line  ? 

Ex.  2.  How  many  straight  lines  are  determined  by  four  points  in 
a plane,  no  three  of  them  being  in  the  same  straight  line  f 

Ex.  3.  If,  in  Fig.  2 above,  A a,  b,  c,  d = 40°,  50°,  60°,  70°  respec- 
tively, find  Z e. 


Ex.  4.  If,  in  Fig.  3 above,  th9  lines  forming  the  angle  q are  per- 
pendicular to  each  other  and  Z p = 47°,  find  the  other  angles  of  the 
figure. 


Ex.  5.  Measure  A a of  Fig.  1 on  preceding  page, 
out  measuring  it.  Now  measure  Z6  and 
compare  the  two  results.  A 

Ex.  6.  Given  QB  ± AB,  PB  ± BC,  and 
AABC=  130°;  find  the  other  angles  of  the 
figure. 


Find  Z b with- 


Ex.  7.  Arrange  five  points  in  a plane 
so  that  the  fewest  number  of  straight  lines  may  pass  through  them, 
no  line  to  pass  through  more  than  three  points. 


PLANE  GEOMETRY 

Book  I 

RECTILINEAR  FIGURES 


Proposition  I.  Theorem 

78.  If  one  straight  line  intersects  another  straight  line , 
J,he  opposite  or  vertical  angles  are  equal. 


Given  the  straight  lines  AB  and  CD  intersecting  at 
the  point  0. 

To  prove  Z AOC  = Z DOB  and  LAOD=  Z COB. 

Proof.  Z A OC+  AAOD=  2 rt.  Z , Art.  73. 

( the  sum  of  two  adjacent  angles  formed  by  one  straight  line  meeting 
another  straight  line  equals  two  right  angles). 

Also  Z.BOD+ l AOI)  = 2 rt.  A , 

( same  teas  on). 

ZAOC+  Z.AOD  = ZBOD+  'AOD,  Ax.  i. 

(things  equal  to  the  same  thing  are  equal  to  each  other). 

Subtracting  ZAOD  from  the  two  equals, 

A AOC  — Z BOD,  Ax.  3. 

(if  equals  be  subtracted  from  equals . the  remainders  are  equal). 

In  like  manner  it  may  be  proved  that 

ZAOD  — Z COB.  Q e-D- 

Ex.  If  in  the  above  figure  /.DOB— 70°,  find  the  other  angles 
without  measuring  them. 


(29) 


30 


BOOK  I.  PLANE  GEOMETRY 


Proposition  II.  Theorem 

79.  If,  from  a point  in  a perpendicular  to  a given  line, 
two  oblique  lines  be  drawn  cutting  off  on  the  given  line  equal 
segments  from  the  foot  of  the  perpendicular , the  oblique  lines 
are  equal  and  make  equal  angles  with  the  perpendicular. 


D 


Given  a line  AB  with  CD  _L  to  it  at  the  point  C,  and 
PR  and  PQ  drawn  from  any  point  as  P in  CD,  cutting  off 
CR  = Q 0 on  AB. 

To  prove  PR  = PQ  and  Z CPR  = Z CPQ. 

Proof.  Fold  over  the  figure  DCB  about  DC  as  an  axis 
till  it  comes  in  the  plane  DCA.  Geom.  Ax.  2. 

Then  ZDCB  — ZDCA  {all  right  A are=).  Art.  72. 

line  CB  will  take  the  direction  of  CA. 

But  CR=CQ.  Hyp. 

point  R will  fall  on  point  Q. 

Hence  line  PR  will  coincide  with  line  PQ,  Art.  66. 

{wily  one  straight  line  can  he  drawn  connecting  two  given  points). 

And  Z CPR  will  coincide  with  Z CPQ. 

.-.  PR  = PQ,  and  Z CPR  = zCPQ,  Art.  47 

{geometric  figures  which  coincide  are  equal). 

Q.  E.  D. 

Ex.  1.  Point  out  the  hypothesis  and  the  conclusion  in  the  general 
enunciation  of  Prop.  I.  Also  point  thorn  out  in  the  particular 
enunciation.  Do  the  same  for  Prop.  II. 

Ex.  2.  If  three  straight  lines  intersect  at  a point,  how  many  of 
the  angles  formed  is  it  necessary  to  measure,  in  order  to  determine 
all  the  angles? 


LINES  AND  ANGLES 


31 


Proposition  III.  Theorem 

80.  From  a given  point  without  a straight  line  but  one 
perpendicular  can  be  drawn  to  the  line. 


Given  the  straight  line  AB , P any  point  without  AB, 
PQ  J_  AB,  and  PB  any  other  line  drawn  from  P to  AB. 

To  prove  that  PB  is  not  J_  AB . 

Proof.  Produce  PQ  to  P'  making  QP'  = PQ.  Draw  BP'. 
Then  BQ  _L  PP'.  Hyp, 

P'Q  = PQ.  Constr. 

/.  BP  = BP'  and  Z PBQ  = / P’BQ,  Art.  79. 
(if,  from  a point  in  a A.  to  a given  line,  two  oblique  lines  be  drawn 
cutting  off  on  the  giv.n  line  equal  segments  from  the  foot  of  the  _L, 
the  oblique  lines  are  equal  and  make  equal  A with  the  J_). 

But  PBP’  is  not  a straight  line,  Art.  66. 
(only  one  straight  line  can  be  drawn  connecting  two  given  points). 

:.  PBP'  is  not  a straight  Z . 

.'.  /.PBQ,  the  half  of  / PBP' , is  not  a right  Z . Ax.  10. 
/.  PB  is  not  _L  AB. 

only  one  perpendicular  can  be  drawn  from  P to  AB. 


Ex.  Three  straight  lines  intersect  at  a point.  Two  of  the  adjaeent 
angles  formed  at  the  point  are  30°  and  40°.  Find  all  the  other  angles 

at  the  point. 


32 


BOOK  I.  PLANE  GEOMETRY 


TRIANGLES 

» 

81.  A triangle  is  a portion  of  a plane  bounded  by  three 
straight  lines,  as  the  triangle  ABC. 

82.  The  sides  of  a triangle  are 
the  lines  which  bound  it;  the 
perimeter  of  a triangle  is  the  sum 
of  the  sides ; the  angles  of  a tri- 
angle are  the  angles  formed  by 
the  sides,  as  the  angles  A,  B 
and  C-,  the  vertices  of  a triangle  are  the  vertices  of  the 
angles  of  the  triangle. 

83.  An  exterior  angle  of  a triangle  is  an  angle  formed  by 
one  side  and  by  another  side 
produced,  as  the  angle  BCD. 

With  reference  to  the  angle 
BCD , the  angles  A and  B 
are  termed  the  opposite  inte- 
rior angles. 

84.  Classification  of  triangles  according  to  relative  length 
of  the  sides.  A scalene  triangle  is  a triangle  in  which  no 
two  sides  are  equal.  An  isosceles  triangle  is  one  in  which 
two  sides  are  equal.  An  equilateral  triangle  is  one  in  which 
all  three  sides  are  equal. 


Scalene 


Isosceles 


Equilateral 


TRIANGLES 


33 


85.  Classification  of  triangles  with  reference  to  character 
of  their  angles.  A right  triangle  is  a triangle  one  of  whose 
angles  is  a right  angle.  An  obtuse  triangle  is  a triangle 
one  of  whose  angles  is  an  obtuse  angle.  An  acute  triangle 
is  a triangle  all  of  whose  angles  are  acute  angles.  An  equi- 
angular triangle  is  one  in  which  all  the  angles  are  equal. 


86.  The  base  of  a triangle  is  the  side  upon  which  the 
triangle  is  supposed  to  stand,  as  AB.  The  angle  opposite 
the  base  is  called  the  vertex  angle,  as 
angle  ACB\  the  vertex  of  a triangle  is 
the  vertex  of  the  vertex  angle  of  the  tri- 
angle. 

The  altitude  of  a triangle  is  the 
perpendicular  from  the  vertex  to  the 
base  or  base  extended,  as  CD. 

87.  In  an  isosceles  triangle,  the  legs  are  the  equal 
sides,  and  the  base  is  the  remaining  side. 

88.  In  a right  triangle,  the  hypotenuse  is  the  side  oppo- 
site the  right  angle,  and  the  legs  are  the  sides  adjacent  to 
the  right  angle. 

89.  Altitudes,  bisectors,  medians.  In  any  triangle,  any 
side  may  be  taken  as  the  base;  hence  the  altitudes  of  a 
triangle  are  the  three  perpendiculars  drawn  one  from  each 
vertex  to  the  side  opposite 


C 


34 


BOOK  I.  PLANE  GEOMETRY 


A bisector  of  an  angle  of  a triangle  is  a line  which.  divides 
this  angle  into  two  equal  parts.  This  bisector  is  usually  pro- 
duced to  meet  the  side  opposite  the  given  angle. 

A median  of  a triangle  is  a ime  drawn  from  a vertex  oi 
the  triangle  to  the  middle  point  of  the  opposite  side.  How 
many  medians  has  a triangle  ? 

90.  Two  mutually  equiangular  triangles  are  triangles 
having  their  corresponding  angles  equal. 

91.  Homologous  angles  of  two  mutually  equiangular 
triangles  are  corresponding  angles  in  those  triangles. 

Homologous  sides  of  two  mutually  equiangular  triangles 
are  sides  opposite  homologous  angles  in  those  triangles. 


We  shall  now  proceed  to  determine  first,  the  properties 
of  a single  triangle,  as  far  as  possible,  then  those  of  two 
triangles. 

92.  Property  of  a triangle  immediately  inferred.  The 

sum  of  any  two  sides  of  a triangle  is  greater  than  the  third 
side.  For  a straight  line  is  the  shortest  line  between  two 
points  (Art.  15.) 


Ex.  1.  Point  out  the  hypothesis  and  conclusion  in  the  general  enun- 
ciation of  Prop.  Ill;  also  point  them  out  in  the  particular  enunciation. 

(As  each  of  the  next  fifteen  Props,  is  studied,  let  the  pupil  do  the 
same  for  it.) 

Ex.  2.  Find  the  angle  whose  complement  is  18°;  whose  supple- 
ment is  76° 

Ex.  3.  If  the  complement  of  an  angle  is  known,  what  is  the 
shortest  way  of  finding  the  supplement  of  the  angle?  If  the  supple- 
ment is  known,  what  is  the  shortest  way  of  finding  the  complement  ? 

Ex.  4.  In  25  minutes,  how  many  degrees  does  the  minute-hand 
of  a clock  travel  ? How  many  does  the  hour-hand  ? 

Ex.  5.  Draw  three  straight  lines  so  that  they  shall  intersect  in 
three  points ; in  two  points ; in  one  point. 


TRIANGLES 


35 


Proposition  IV.  Theorem 

93.  Any  side  of  a triangle  is  greater  than  the  difference 
between  the  other  tivo  sides. 


B 


Given  AB  any  side  of  the  A ABC,  and  AC  >BC. 

To  prove  AB  >AC — BC. 

Proof.  AB  + BC  > AC,  Art.  92. 

(the  sum  of  any  two  sides  of  a triangle  is  greater  than  the  third  side). 

Subtracting  BC  from  each  member  of  the  inequality, 

AB>AC—  BC,  As.  9, 

{ if  equals  he  subtracted  from  unequals,  the  remainders  are  unequal  in  the 
same  order).  q.  e.  D. 

94.  Cor.  The  perpendicular  is  the  shortest  line  that 
can  be  drawn  from  a given  point  to  a given  line. 

For,  in  the  Fig.  page  31, 

PP'  < PR  + BP',  Art  92. 

Or,  2 PQ  <2  PR.  Ax.  8. 

PQ  <PB.  Ax.  10. 

Hence,  Def.  The  distance  from  a point  to  a line  is  the 
perpendicular  drawn  from  the  point  to  the  line. 

Ex.  1.  If  one  side  of  an  equilateral  triangle  is  4 inches,  what  is  its 
perimeter  ? 

Ex.  2.  Is  it  possible  to  form  a triangle  whose  sides  are  6,  9 and 
17  inches  f Try  to  do  this  with  the  compasses  and  ruler. 

Ex.  3.  Is  it  possible  to  form  a triangle  in  which  one  side  is  10 
inches  and  the  difference  of  the  other  two  sides  is  12  inches  ? 

Ex.  4.  On  a given  line  as  base,  by  exact  use  of  ruler  and  com- 
passes, construct  an  equilateral  triangle. 


36 


BOOK  I.  PLANE  GEOMETBY 


Proposition  V.  Theorem 

95.  If,  from  a point  within  a triangle , two  lines  oe  drawn 
to  the  extremities  of  one  side  of  the  triangle,  the  sum  of 
the  other  tivo  sides  of  the  triangle  is  greater  than  the  sum 
of  the  two  lines  so  drawn. 


B 


Given  P any  point  within  the  triangle  ABC,  and 
PA  and  PC  lines  drawn  from  P to  the  extremities  of  the 
side  AC. 

To  prove  AB  + BC  > AP  + PC. 

Proof.  Produce  the  line  AP  to  meet  BC  at  Q. 

Then  AB  + BQ  > AP  + PQ,  Art.  15. 

(a  straight  line  is  the  shortest  line  connecting  two  points) . 

Also  PQ  + QC  > PC, 

{same  reason). 

Adding  these  inequalities, 

AB  + BQ  + PQ  + QC  > AP  + PQ  + PC.  Ax.  9. 
Substituting  BC  for  its  equal  BQ  + QC, 

AB  + BC  + PQ  > AP  + PQ  + PC.  Ax.  8. 
Subtracting  PQ  from  each  side, 

AB  + BC  > AP  + PC.  A*-  9- 

Q.  E.  D. 


Ex.  On  a given  line  as  base,  by  exact  use  of  ruler  and  compasses, 
construct  an  isosceles  triangle  each  of  whose  legs  is  double  the  base. 


TRIANGLES 


37 


Proposition  VI.  Theorem 

96.  Two  triangles  are  equal  if  two  sides  and  the  in- 
cluded angle  of  one  are  equal , respectively,  to  two  sides  and 
the  included  angle  of  the  other. 

B E 


Given  the  triangles  ABC  and  BEE  in  which  AB  — BE, 
AC=BF,  and  £A=  ID. 

To  prove  A ABC  — A BEE. 

Proof.  Place  the  A ABC  upon  the  A DEF  so  that  the 
line  AC  eoncides  with  its  equal  BE.  Geom.  Ax.  2. 

Then  the  line  AB  will  take  the  direction  of  DE, 

(for  ZA  = ZD  by  hyp.) . 

Also  the  point  B will  fall  on  E, 

( for  line  AB  = line  DE  by  liyp.). 

Hence  the  line  BG  will  coincide  with  the  line  EE,  Art.  66. 
(only  one  straight  line  can  be  drawn  connecting  two  points). 

A ABC  and  BEE  coincide. 

A A ABC  = A DEF,  Art.  47, 

(geometric  figures  which  coincide  are  equal) . 

Q.  E.  D. 

Ex.  1.  What  kind  of  proof  is  used  in  Prop.  VI  9 (See  Art.  62). 

Ex.  2.  If  A A,  B and  C=60°,  70°,  50°,  AB=  16,  AC=  19,  BC=  18: 
also  Z 13=60°,  DE=  16,  73^=19 : find  Z E and  F and  side  EE 
without  measuring  them 


38 


BOOK  I.  PLANE  GEOMETRY 


Proposition  VII.  Theorem 

97.  Two  triangles  are  equal  if  two  angles  and  the  in- 
cluded side  of  one  are  equal , respectively , to  two  angles  and 
the  included  side  of  the  other. 

B E 


Given  the  A ABC  and  DBF  in  which  Zi=  ZD, 
ZC  = IF,  and  AC—DF. 

To  prove  A ABC—  A DEF. 

Proof.  Place  the  A ABC  upon  the  A DEF  so  that  AC 
shall  coincide  with  its  equal  DF.  Geom.  Ax.  2. 

Then  AB  will  take  the  direction  of  DF, 

( for  A A = ID  by  hyp.), 

and  the  point  B will  fall  somewhere  on  the  line  DE  or  DE 
produced. 

Also  the  line  CB  will  take  the  direction  of  FE, 

{for  ZC  = AFby  hyp.), 

and  the  point  B will  fall  on  FF  or  FE  produced. 

.*.  point  B falls  on  point  E,  Art.  64. 

( two  straight  lines  can  intersect  in  but  one  point). 

A ABC  and  DEF  coincide. 

.-.  A ABC=  ADEF,  Art.  47 

{geometric  figures  which  coincide  are  equal).  q e j)_ 

Ex.  1.  What  kind  of  proof  is  used  in  Prop.  VTI.  f 
Ex.  2.  If  A A,  B,  C= 65°,  55°,  60°,  AB=  24,  AC=\8,  BC=Z7, 
also  A D,  F=  65°,  60°,  and  ZtF=18;  find  DE,  EF,  and  AE. 

Ex.  3.  Construct  by  exact  use  of  ruler  and  compasses  a scalene 
triangle  whose  sides  are  2,  3 and  4 times  a given  line. 


TRIANGLES 


39 


Proposition  YIII.  Theorem 


98.  Two  right  triangles  are  equal  if  the  hypotenuse  and 
an  acute  angle  of  one  are  equal  to  the  hypotenuse  and  an 
acute  angle  of  the  other. 


Given  the  right  A ABC  and  DEF  in  which  hypote- 
nuse AB  = hypotenuse  DE,  and  Z A = Z D. 

To  prove  A ABC  = A DEF. 

Proof.  Place  the  A ABC  upon  the  A DEF  so  that  the 
side  AB  shall  coincide  with  its  equal,  the  side  DE , the 
point  A coinciding  with  the  point  D. 

Then  the  line  AC  will  take  the  direction  of  DF, 

( for  Z A = Z D by  htjp.) 

Also  the  side  BC  will  coincide  with  the  side  EF,  Art.  80. 
( from  a given  point,  E,  without  a straight  line,  DF,  hut  one  _L  can  he 
drawn  to  the  line). 

A ABC  and  DEF  coincide. 

A ABC=  A DEF,  Art.  47. 

{geometric  figures  which  coincide  are  equal) . 

Q.  E.  D. 

Ex.  Construct  exactly  an  equilateral  triangle,  each  of  whose  sides 
ohall  be  double  a given  line. 


40 


BOOK  I,  PLANE  GEOMETRY 


Proposition  IX.  Theorem 


99o  In  an  isosceles  triangle  the  angles  opposite  the  equal 
Aides  are  equal. 


B 


Given  the  isosceles  A ABC  in  which  AB  = BC. 
To  prove  Z A = Z C. 

Proof.  Let  BD  be  drawn  so  as  to  bisect  /.ABC. 


Then,  in  the  A ABB  and  BBC, 

AB  — BG.  Hyp. 

Also  BB  = BB,  Ident. 

And  Z ABB=  Z CBB.  Constr. 

.•.  A ABB  — A CBB , Art.  96. 


{two  A are  equal  if  two  sides  and  the  included  /_  of  one  are  equal , respec- 
tively, to  two  sides  and  the  included  Z.  of  the  other.) 

:.  ZA=/C, 

{homologous  A of  equal  A). 


Ex.  1.  On  a given  line  as  base,  construct  exactly  an  equilateral 
triangle  above  the  line  and  another  below  it. 

Ex.  2.  On  a given  line  as  base,  construct  exactly  an  isosceles 
triangle  whose  leg  shall  be  equal  to  a given  line;  make  the  same  con- 
struction below  the  given  line  and  join  the  vertices  of  the  two 
isosceles  triangles. 


TRIANGLES 


41 


Proposition  X.  Theorem  (Converse  of  Prop.  IX) 

100.  If  two  angles  of  a triangle  are  equal , the  sides 
opposite  are  equal,  and  the  triangle  is  isosceles. 


B 


Given  the  triangle  ABC  in  which  Z1  = ZBCA. 
To  prove  AB—BC. 


Proof. 

If  the  sides  AB  and  BG  are 

not  equal,  one  of 

them  must 
than  BG. 

be  longer  than  the  other. 

Let  AB  be  longer 

On  AB  mark  off  AB  — BG  Draw  DC. 
Then,  in  the  A ABC  and  ADC, 

AB  = BG, 

Constr. 

AC— AC, 

Ident. 

ZDAG=  ZBGA. 

Hyp. 

:.  A DAG=  A BGA, 

Art.  90. 

( two  A are  equal  if  two  sides  and  the  included  Z of  one  are  equal, 
respectively,  to  two  sides  and  the  included  Z of  the  other). 


Or  a part  is  equal  to  the  whole,  which  is  impossible.  Ax.  7. 
Hence  AB  cannot  be  greater  than  BG. 

In  like  manner  it  may  be  shown  that  AB  is  not  less 
than  BG. 

Hence  AB—BG. 


Ex.  1.  What  method  of  proof  is  used  in  Prop.  X? 

Ex.  2.  If  in  a triangle  DEF,  ZD  = 36°,  ZE= 36°  and  DF=  12, 
find  EF,  Draw  a figure  and  on  it  mark  the  value  of  the  parts  named. 


42 


BOOK  I.  PLANE  GEOMETEY 


Proposition  XI.  Theorem 


101.  Two  triangles  are  equal  if  the  three  sides  of  one 
are  equal,  respectively , to  the  three  sides  of  the  other. 


Given  the  A ABC  and  BEE  in  which  AB  — BE,  BG= 
EE,  and  AC=DF. 

To  prove  A ABC=  A BEE. 


Proof.  Place  the  A ABC  so  that  its  longest  side  AC 
shall  coincide  with  its  equal  BE  in  the  A BEE  and  the 
vertex  B shall  fall  on  the  opposite  side  of  BE  from  E. 

. Geom.  Ax.  2. 

Draw  the  line  EB. 

Then  BE=BB  (Hyp.)  .*.  A BEB  is  isosceles.  Def. 

.".  Z p = Z r , Art.  99. 

(in  an  isosceles  A the  A opposite  the  equal  sides  are  equal). 

In  like  manner,  in  the  A BEE,  Z q — Zs. 

Adding, 

ZpfrZq—Zr-fZs,  Ax.  2. 

Or  ZBEF=ZBBF.  Ax.  6. 

.*.  A BEE  = A BBF,  Art.  96. 

(two  A are  equal  if  two  sides  and  the  included  Z of  one  are  equal, 
respectively , to  two  sides  and  the  included  Z of  the  other ) . 

.*.  A ABC  — A BEE.  Ax.  l. 


Ex.  1.  Construct  two  equilateral  triangles  on  the 
same  base,  one  above  and  the  other  below,  and  join  the 
two  vertices.  Prove  that  the  line  joining  the  vertices 
bisects  the  vertex  angles,  and  also  bisects  the  base  at 
right  angles. 

Ex.  2.  Hence,  at  any  point  in  a given  straight  line, 
construct  exactly  by  use  of  ruler  and  compasses  a 
perpendicular  to  that  line. 


Q.  E.  D. 


TRIANGLES 


43 


Proposition  XII.  Theorem  "> 

102.  Two  right  triangles  are  equal  if  the  hypotenuse  and 
a leg  of  one  are  equal  to  the  hypotenuse  and  a leg  of  the  other. 


E 


Given  two  right  A ABC  and  BEF  having  the  hypote* 
nuse  AB  = hypotenuse  BE , and  BG—EF. 

To  prove  A ABG=  A BEE. 

Proof.  Place  the  A ABC  so  that  BG  shall  coincide  with 
its  equal,  EE,  and  A fall  on  the  opposite  side  of  EE  from 
B , at  A' . Geom.  Ax.  2. 

Then  A'E  and  FB  will  form  a straight  line,  A'FB,  Art.  74. 
(if  two  adj.  A are  together  equal  to  two  rt.  A , their  ext.  sides  form  one 
and  the  same  straight  line) . 

But  A'E=EB.  Hyp. 

.*.  A A'EB  is  isosceles.  Def. 

ZA'=  ZB,  Art.  99. 

(in  an  isosceles  A the  A opposite  the  equal  sides  are  equal. ) 

A A'EF=  A BEF,  Art.  98. 

( two  right  A are  eqiuxl  if  the  hypotenuse  and  an  acute  Z of  one  are  equal 
to  the  hypotenuse  and  an  acute  Z of  the  other). 

Or  A ABC=  A BEF.  Ax.  1. 

Q.  E.  D. 

Ex.  1.  By  making  the  same  construction 
as  in  Ex.  1,  p.  42,  bisect  any  given  straight 
line.  q 

Ex.  2.  Bisect  any  given  angle  AOB , 


44 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XIII.  Theorem 

103.  An  exterior  angle  of  a triangle  is  greater  than 
either  opposite  interior  angle. 

B F 

■D 

Given  ZBCD  an  exterior  Z of  the  A ABC. 

To  prove  ZBCD  greater  than  ZABC  or  ZBAG. 

Proof.  Let  E be  the  middle  point  of  the  line  BO. 

Draw  AE  and  produce  it  to  F,  making  FE=AE.  Draw 
EG. 

Then,  in  the  A AEB  and  EEC, 

AE  = FE,  and  BE  = CE,  Constr. 

ZBEA=  Z EEC  {being  vertical  A).  Art.  78. 

.*.  A AEB  = A EEC,  Art.  96. 

( two  A are  equal  if  two  sides  and  the  included  Z of  one  are  equal,  re- 
spectively, to  two  sides  and  the  included  Z of  the  other). 

:.  ZABE  = Z FCE, 

( being  homologous  A of  equal  A ) . 

But  ZBCD  is  greater  than  Z FCE,  Ax.  7. 

( the  whole  is  greater  than  any  of  its  parts). 

Substituting  Z ABE  for  its  equal  Z FCE,  Ax.  8. 

ZBCD  is  greater  than  ZABE,  that  is,  than  ZABC. 

Similarly,  by  drawing  a line  from  B through  the  midpoint 
of  AC  and  by  producing  BC  through  C to  a jjoint  H,  it  may 
be  shown  that  Z ACE  {—  ZBCD)  is  greater  than  / BAC. . 

Q.  E.  D. 

Ex.  On  a given  line,  as  base,  construct  exactly  an  isosceles  triangle 
each  of  whose  legs  equals  half  a given  line. 


TRIANGLES 


45 


Proposition  XIV.  Theorem 

104.  If  two  sides  of  a triangle  are  unequal , the  angles 
spposite  are  unequal , and  the  greater  angle  is  opposite  the 
greater  side 

B 


Given  the  side  BC  > side  AB  in  the  A ABC. 

To  prove  ABAC  greater  than  Z C. 

Proof.  On  the  side  BC  take  BD  equal  to  AB  and  draw 
AD. 

Then,  in  the  isosceles  A ABD,  Zr  — Zs,  Art.  99. 

(in  an  isosceles  A the  A s opposite  the  equal  sides  are  equal ) . 

Z BAC  is  greater  than  Zr,  Ax.  7. 

(the  whole  is  greater  than  any  of  its  parts) . 

:.  ABAC  is  greater  than  Zs.  Ax.  8. 

But  Zs  is  an  exterior  Z of  the  A ADC. 

:.  Zs  is  greater  than  Z C,  Art.  103 

(aw  ext.  Z of  a A is  greater  than  either  oppositeint.  Z ). 

Much  more,  then,  is  Z.BAC  {which  is  greater  than  Zs) 
greater  than  Z C,  Ax.  12. 

(if,  of  three  quantities,  the  first  is  greater  than  the  second,  and  the  second 
is  greater  than  the  third,  then  the  first  is  greater  than  the  third) . 

Q.  E.  D. 

105.  Note.  The  essential  steps  of  the  above  proof  may  be 
arranged  in  a single  statement,  thus: 

ABAC  > Ar  = Zs  > Z C ABAC  is  greater  than  Z C. 


Ex.  1.  Which  is  the  longest  side  of  a right  triangle  ? of  an  obtuse 
triangle  ? 

Ex.  2.  Construct  exactly  an  equilateral  triangle,  each  of  whose 
sides  is  half  a given  line. 


46 


BOOK  I.  PLANE  GEOMETKY 


Proposition  XV.  Theorem  (Converse  of  Prop.  XIV) 

106.  If  two  angles  of  a triangle  are  unequal , the  sides 
opposite  are  unequal , and  the  greater  side  is  opposite  the 
greater  angle. 


Given  Z A greater  than  Z C in  the  A ABC. 

To  prove  BC  > AB. 

Proof.  BC  either  equals  AB,  or  is  less  than  AB,  or  is 
greater  than  AB. 

But  BC  cannot  equal  AB, 

for,  if  it  did,  Z1  would  equal  AC,  Art.  &9. 

( being  opposite  equal  sides  in  an  isosceles  A). 

But  this  is  contrary  to  the  hypothesis. 

Also  BC  cannot  be  less  than  AB, 

for,  if  it  were,  Z A would  be  less  than  Z C,  Art.  104. 
(if  two  sides  of  a A are  unequal,  the  A opposite  are  unequal,  and  the 
greater  A is  opposite  the  greater  side). 

This  is  also  contrary  to  the  hypothesis. 

A BC>AB, 

(for  it  neither  equals  AB,  nor  is  less  than  AB). 

Q.  E.  B. 


Ex.  1.  Draw  a triangle  the  altitude  of  which  falls  on  the  base 
produced.  What  kind  of  a triangle  is  this  ? 


Ex.  2.  Draw  a triangle  the  altitude  of 
which  coincides  with  one  side.  What  kind  of 
a triangle  is  this  ? 


Ex.  3.  By  exact  use  of  the  ruler  and  com- 
passes, draw  a perpendicular  to  a given  line 
from  a given  point  without  the  line, 


TRIANGLES 


47 


Proposition  XVI.  Theorem 

107.  If  two  triangles  have  two  sides  of  one  equal , respec- 
tively, to  two  sides  of  the  other,  hut  the  included  angle  of 
the  first  greater  than  the  included  angle  of  the  second,  then 
the  third  side  of  the  first  is  greater  than  the  third  side  oj 
the  second. 

B E 


Given  the  A ABC  and  DBF  in  which  AB  = DB, 
BC=BF,  and  /.ABC  is  greater  than  /E. 

To  prove  AC  > DF. 

Proof.  Place  the  A DEF  so  that  the  side  DE  coincides 
with  its  equal,  the  side  AB,  and  F takes  the  position  F'. 

Geom.  Ax.  2. 

Let  the  line  BE  bisect  the  /F'BC  and  meet  the  line 
AC  at  H.  Draw  F'H. 

Then,  in  the  A F'BH  and  BE C,  F'B  = BC,  Hyp. 


BE —BE,  Ident. 

Z F'BE=  Z CBE.  Constr. 

A F'BE=  A EEC.  (Why?) 

F'E  = GE,  {homologous  sides  of  equal  A) . 

But  AE  + EF'  > AF' , Art.  92 

( the  sum  of  any  two  sides  of  a A is  greater  than  the  third  side). 

Substituting  for  EFr  its  equal  EC, 

AE  -f  EC,  or  AC  > AF'.  Ax.  8. 

AC>  DF.  Ax.  8. 

Q.  E.  D. 


Ex.  1.  Draw  a figure  for  Prop,  XVI  in  which  the  sides  and  an 
gles  are  of  such  a size  that  F falls  within  the  triangle  ABC. 

Ex.  2.  Draw  another  figure  in  which  F falls  on  the  side  AC, 


48 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XVII.  Theorem  (Converse  of  Prop.  XVI) 


108.  If  two  sides  of  a triangle  are  equal , respectively , 
to  two  sides  of  another  triangle,  but  the  third  side  of  the 
first  is  greater  than  the  third  side  of  the  second,  then  the 
angle  opposite  the  third  side  of  the  first  triangle  is  greater 
than  the  angle  opposite  the  thud  side  of  the  second. 


Given  the  A ABC  and  DBF  having  AB—DE,  BC= 
EF,  but  AC  > DF. 

To  prove  Z B greater  than  Z E. 

Proof.  The  Z B either  equals  Z E,  or  is  less  than  Z E, 
or  is  greater  than  Z E. 

But  Z B does  not  equal  Z E, 

for,  if  it  did,  A ABC  would  = A DEE,  Art.  96. 

( two  A are  equal  if  two  sides  and  the  included  Z of  one  are  equal, 
respectively,  to  two  sides  and  the  included  Z of  the  other), 

and.  AC  would  equal  DF  {homologous  sides  of  equal  A), 
which  is  contrary  to  the  hypothesis. 

Also  if  ZB  were  less  than  ZE, 

side  AC  would  be  less  than  side  DF,  Art.  107. 
{if  two  A.  have  two  sides  of  one  equal,  respectively,  to  two  sides  of  the  other , 
but  the  included  Z of  the  first  greater  than  the  included  Z of 
the  second,  then  the  third  side  of  the  first  is  greater 
than  the  third  side  of  the  second ) . 

But  this  is  also  contrary  to  the  hypothesis. 

Hence  Z B is  greater  than  Z E, 

( for  it  neither  equals  Z E,  nor  is  less  than  Z.E). 

Q.  E.  D. 


Ex.  On  a given  line  («)  con- 
struct a triangle  whose  other  two  sides 

are  equal  to  two  given  ’ines  Im  and  n ). 


LINES 


49 


PROPERTIES  OF  LINES  PROVED  BY  USE  OF  TRIANGLES 

Proposition  XVIII.  Theorem 

109.  Of  lines  drawn  from  the  same  point  in  a perpen- 
dicular and  cutting  off  unequal  segments  from  the  foot  of 
the  perpendicular,  the  more  remote  is  the  greater. 


P 


Given  PO  ± AB,  PT  and  PQ  oblique  to  AB,  and 
OT  > OQ. 

To  prove  PT  > PQ. 

Proof.  Produce  PO  to  the  point  P'  making  OP'=OP. 
On  AB  take  OR = OQ.  Draw  PR,  P'R,  P'T. 

Then  PQ  = PR,  Art.  79. 

{if  from  a point  in  a L to  a given  line,  tico  oblique  lines  be  drawn,  cutting 
off  on  the  given  line  equal  segments  from  the  foot  of  the  -L,  the 
oblique  lines  are  equal). 

In  A PTP' , PT  + TP'  > PR  + RP> , Art.  95. 

{if,  from  a point  within  a A,  two  lines  be  drawn  to  the  extremities  of  a 
side  of  the  A,  the  sum  of  the  other  two  sides  of  the  A is  greater 
than  the  sum  of  the  two  lines  so  drawn). 

Bui  OTis  _L  PP'  and  PT  and  P'T  cut  off  equal  segments, 
PO  and  P'0,  from  the  foot  of  the  JL  AO. 

Hence  PT  = P'T.  In  like  manner  PR  = P'R.  Art.  79. 
Hence,  by  substitution,  2 PT  > 2 PR . Ax.  8. 

.*.  PT  > PR.  Ax.  10. 

PT  > PQ.  Ax.  8. 

Q.  E.  X). 


D 


50 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XIX.  Theorem  (Converse  op  Prop.  II) 

110.  Equal  oblique  lines  drawn  from  a point  in  a per- 
pendicular cut  off  equal  segments  from  the  foot  of  the 
perpendicular. 


Given  PC  J.  AB,  PR  and  PT  oblique  to  AB,  and 
PR  = PT. 

To  prove  CR—CT. 

Proof.  In  the  right  A RPC  and  CPT, 

PC=  PC.  Ident. 

Also  PR  — PT.  Hyp. 

A RPC=  A CPT,  Art.  102. 

( two  right  A are  equal  if  the  hypotenuse  and  a leg  of  one  are  equal  to  the 
hypotenuse  and  a leg  of  the  other). 

:.  RC=CT, 

( homologous  sides  of  equal  A). 

Q.  E.  J>. 

111.  Cor.  Of  two  unequal  lines  drawn  from  a point 
in  a perpendicular , the  greater  line  cuts  off  the  greater 
segment  from  the  foot  of  the  perpendicular . 

Thus,  if  PT  > PQ  (Fig.  of  Prop.  XVIII),  OT  cannot 
= OQ  (Art.  79);  nor  is  OT < OQ  (Art.  109)  .■.  OT>  OQ. 

Hence,  also,  from  a given  point  only  two  equal  straight 
lines  can  be  drawn  to  a given  line. 


LINES 


51 


Proposition  XX.  Theorem 

112.  I.  Every  point  in  the  perpendicular  bisector  of  a 
line  is  equally  distant  from  the  extremities  of  the  line;  and 
II.  Every  point  not  in  the  perpendicular  bisector  is 
unequally  distant  from  the  extremities  of  the  line. 


Given  0 the  middle  point  of  the  line  AB,  OC  _L  AB, 
P any  point  in  OC,  and  Q any  point  not  in  OC. 

To  prove  AP=PB,  but  QA  and  QB  unequal. 

Proof.  I.  AP=PB,  Art.  79. 

< if,  from  a point  in  a ± to  a given  line,  two  oblique  lines  be  drawn 
cutting  off  on  the  given  line  equal  segments,  etc.). 

II.  Since  Q is  not  in  the  line  OC,  either  AQ  or  QB  must 
cut  the  line  OC. 

Let  AQ  intersect  OC  in  the  point  R and  join  RB. 

Then  AR  — RB,  {by  first  part  of  this  theorem). 

To  each  of  these  equals  add  RQ. 

Then  AR  RQ=RB  -f-  RQ.  Ax.  2. 

But  RB  + RQ  > QB.  ,Why?) 

.\  by  substitution,  AR  + RQ,  or  AQ  > QB.  Ax.  8. 

Q.  E.  D. 

113.  Cor.  Two  points  each  equidistant  from  the  ex- 
tremities of  a line  determine  the  perpendicular  bisector  of 
the  line. 

This  corollary  gives  a useful  method  of  determining  the 
perpendicular  bisector  of  a given  straight  line,  by  determin- 
ing two  points  only  of  the  perpendicular  bisector. 


52 


BOOK  I.  PLANE  GEOMETKY 


LOCI 

114.  Def.  The  locus  of  a point  is  the  path  of  a point 
moving  according  to  a given  geometric  law. 

Thus,  if  a point  move  in  a plane  so  as  to  he  always  two  inches  dis- 
tant from  a given  point,  its  locus  is  the  circumference  of  a circle 
whose  center  is  the  given  point,  and  whose  radius  is  a line  two  inches 
in  length. 

Thus,  also,  the  locus  of  a point  moving  so  as  to  be  equidistant 
from  two  given  parallel  lines  is  a straight  line  lying  midway  between 
the  two  given  lines. 

The  locus  of  a point  may  consist  of  two  or  more  sepa- 
rate lines  or  parts. 

Thus,  the  locus  of  a point  moving  so  as  to  be  always  at  a given 
distance  from  a given  line  is  two  lines,  one  on  either  side  of  the 
given  line,  at  the  given  distance  from  it. 

1 15.  ' Demonstration  of  loci.  In  order  to  prove  that  a 
given  line  is  the  locus  of  a giveu  point  moving  according 
to  a given  geometric  law,  it  is  necessary: 

1.  To  prove  that  every  point  in  the  given  line  satisfies  the 
given  law  or  condition. 

2.  To  prove  that  every  point  not  in  the  given  line  does  not 
satisfy  the  given  law  or  condition. 

Instead  of  2,  it  may  be  proved  that  every  point  which  satisfies  the 
given  condition  lies  in  the  given  line. 

Hence,  in  Prop.  XX  it  has  been  proved  that  the  per - 
pendicular  bisector  of  a line  is  the  locus  cf  all  points  equi- 
distant from  the  extremities  of  the  line. 

116.  Use  of  loci.  Loci  are  useful  in  determining  a 
point  (or  points)  which  shall  satisfy  two  or  more  geomet- 


LOCI 


53 


rical  conditions.  For,  by  finding  the  locns  of  all  points 
which  satisfy  one  of  the  given  conditions,  and  also  finding 
the  locus  of  all  points  which  satisfy  a second  condition, 
and  then  finding  the  intersection  of  these  two  loci,  we 
obtain  the  point  (or  points)  which  satisfy  both  conditions 
at  the  same  time 

Thus,  if  it  be  required  to  find  the  points  which  are  two  inches 
from  one  given  point  and  three  inches  from  another  given  point,  the 
two  given  points  being  four  inches  apart,  the  required  points  are  the 
intersections  of  the  circumferences  of  two  circles. 

Let  the  pupil  make  a construction  and  obtain  the  required  points. 


Ex.  1.  Draw  the  locus  of  a point  moving  at  the  distance  of  one 
inch  from  a given  point. 

Ex.  2.  Draw  exactly  the  locus  of  a point  moving  at  a distance  of 
one  inch  from  a given  line. 

Ex.  3.  Draw  exactly  the  locus  of  a point  moving  so  as  to  be  equi- 
distant from  the  extremities  of  a given  line  one  inch  long. 

Ex  4.  How  many  points  in  a plane  are  necessary  to  determine 
two  parallel  lines  ? Three  parallel  lines  ? (See  Art.  47.) 

Ex.  5.  Are  two  triangles  equal  if  three  angles  of  one  equal  the 
corresponding  three  angles  of  the  other  ? Illustrate  by  drawing  a 
figure. 

Ex.  6.  Draw  three  isosceles  triangles  on  the  same  base  and 
^connect  their  vertices.  What  truth  is  illustrated  by  this  figure? 

Ex.  7.  Draw  a straight  line  and  locate  a point  2 inches  from  it. 
By  the  use  of  loci,  locate  the  points  which  are  IX  inches  from  the 
given  line  and  at  the  same  distance  from  the  given  point. 


54 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXI.  Theorem 

117.  I.  Every  point  in  the  bisector  of  an  angle  is  equi- 
distant from  the  sides  of  the  angle;  and 

II.  Conversely,  every  point  equidistant  from  the  side, 
of  an  angle  lies  in  the  bisector  of  the  angle. 


I.  Given  PB  the  bisector  of  the  angle  ABC,  P any 
point  in  PB,  PQ  _L  BA,  and  PR  _L  BC. 


To  prove 

PQ=PR. 

Proof. 

In  the  rt.  A PBQ  and  PBR, 

PB  = PB. 

Ideat. 

Also 

/PBQ=  / PBR. 

Hyp. 

:.  APBQ=APBR, 

Art.  98. 

( two  rt.  A are  equal  if  the  hypotenuse  and  an  acute  Z of  one,  etc.). 

Hence  PQ=PR, 

( homologous  sides  of  equal  A ) . 


II.  Given  /.ABC,  PQAAB,  PR  ABC,  and  PQ  = PR. 
To  prove  that  PB  is  the  bisector  of  /ABC. 


Proof. 

In  the  right  A PBQ  and  PBR , 

PB  = PB. 

Ident. 

Also 

PQ  = PR. 

Hyp. 

LOCI 


55 


.*.  A PBQ=  APBR,  Art.  102. 

(two  rt.  A are  equal  if  the  hypotenuse  and  a leg  of  one,  etc.) . 

.'.  / ABP—  / CBP  ( homologous  A of  equal  A). 

Or  /.ABC  is  bisected  bv  BP. 

Q.  E.  D. 

9 

118.  Cor.  In  Prop.  XXI  it  has  been  proved  that  the 
bisector  of  an  angle  is  the  locus  of  all  points  equidistant  from 
the  sides  of  the  angle,  for  it  has  been  proved  that  every 
point  in  the  given  line  satisfies  the  given  law  or  condition, 
and  that  every  point  which  satisfies  the  given  condition 
lies  in  the  given  line  (see  Art.  115). 

119.  Def.  A transversal  is  a 
line  that  intersects  two  or  more 
other  lines.  Thus,  EF  is  a trans- 
versal of  the  l’nes  AB  and  CD. 

If  two  lines  are  cut  by  a trans- 
versal, it  is  convenient  to  give  the 
eight  angles  of  intersection  special 
names. 

a,  b,  g,  h,  are  called  exterior  angles. 

c,  d,  e,  f,  are  called  interior  angles. 

c,  f,  form  a pair  of  alternate-interior  angles. 

b,  f,  form  a pair  of  exterior-interior  angles  on  the 
same  side  of  the  transversal. 

Let  the  pupil  name  another  pair  of  alternate-interior 
angles;  also  name  another  pair  of  exterior- interior 
angles  on  the  same  side  of  the  transvei’sal;  also  name  a 
pair  of  interior  angles  on  the  same  side  of  the  trans- 
versal, 


F 


56 


BOOK  I.  PLANE  GEOMETRY 


PARALLEL  LINES 

120.  Def.  Parallel  lines  have  already  been  defined 
(Art.  41)  as  straight  lines  which  lie  in  tne  same  plane  ana  ao 
not  meet,  however  far  they  be  produced. 

What  is  the  fundamental  axiom  concerning  parallel 
lines?  (see  Art.  47.) 


Proposition  XXII.  Theorem 


121.  Two  straight  lines  in  the  same  plane , perpendicular 
to  the  same  straight  line,  are  parallel , 


P 
A - 


-B 


C- 

Q 


-D 


Given  the  lines  AB  and  CD  J_  line  PQ  and  in  the 
same  plane. 

To  prove  AB  ||  CD. 

Proof.  If  AB  and  CD  are  not  parallel,  they  will  meet 
if  sufficiently  produced.  Art.  120. 

We  shall  then  have  two  _k  from  the  same  point  to  the 
line  PQ. 

But  this  is  impossible,  Art.  80 

{from  a given  point  without  a straight  line  but  one  J.  can  be  drawn  to 

the  line). 

Hence  AB  and  CD  never  meet. 

AB  and  CD  are  parallel.  Def. 

Q.  £.  D. 


PARALLEL  LINES 


57 


122.  Cor.  Tivo  straight  lines  parallel  to  a third  straight 
line  are  parallel  to  each  other; 

Lines  parallel f to  parallel  lines  are  parallel; 

Lines  perpendicular  to  parallel  lines  are  parallel; 

Lines  perpendicular  to  non-parallel  lines  are  not  parallel. 


Proposition  XXIII.  Theorem 

123.  If  a straight  line  is  perpendicular  to  one  of  two 
given  parallel  lines  it  is  perpendicular  to  the  other  also. 


Given  AB  ||  CD,  and  PQ  _L  AB. 

To  prove  PQ  _L  CD. 

Proof.  Let  CP  be  drawn  J_  PQ  at  G. 

Then  CF  ||  AB,  Art.  121. 

( two  straight  lines  in  the  same  plane  JL  same  straight  line  are  ||). 

But  CD  ||  AB.  Hyp. 

.•.  CP  coincides  with  CD,  Geom  ,Ax.3. 
( through  a given  point  one  straight,  line,  and  only  one,  can  he  drawn 
||  another  given  straight  line). 

But  PQ  _L  CF.  Constr. 

Hence  PQ  J_  CD, 

(for  CL  coincides  with  CF,  to  which  PQ  is  ±). 

0,  S*  D9 


58 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXIV.  Theorem 

124.  If  two  parallel  straight  lines  are  cut  by  a trans- 
versal, the  alternate  interior  angles  are  equal. 


Given  the  1 1 lines  AB  and  CD  cut  by  the  transversal 
PQ  at  the  points  G and  E respectively. 

To  prove  /.AGE  = Z GED. 

Proof.  Through  B,  the  middle  point  of  EG,  let  tha  line 
HE  be  drawn  _L  AB. 

Then  HE  JL  CD,  Art.  123. 

{if  a straight  line  is  J_  one  of  two  ||  lines,  it  is  i.  the  other  also) . 

, In  the  right  A GRH  and  ERF, 

GR  = ER,  Oonstr. 

Z GRH  = / ERF.  (Why!) 

.'.  A GRIT  = A ERF,  Art.  98. 

( two  right  A are  equal  if  the  hypotenuse  and  an  acute  Z of  one  — 
the  hypot.  and  an  acute  z of  the  other). 

/ HGR  = Z REF,  or,  /AGE  = / GED, 
(homologous  A of  equal  A). 

Q.  E.  D. 


Bx.  In  the  above  figure  let  the  pupil  show  that  Z BGE  = / GEC . 


PARALLEL  LINES 


59 


Prop.  XXV.  Theorem  (Converse  of  Prop.  XXIV) 

125.  If  two  straight  lines  are  cut  by  a transversal, 
malting  the  alternate  interior  angles  equal,  the  two  straight 
lines  are  parallel. 


Given  the  two  lines  AB  and  CD  cut  by  the  transversal 
PQ  at  the  points  F and  G,  making  Z.AFG  — Z FGD . 

To  prove  AB  ||  CD. 

Proof.  Through  F let  the  line  KL  be  drawn  ||  CD. 

Then  ZKFG  = A FGD,  Art.  124. 

(if  two  ||  st.  lines  are  cut  by  a transversal,  the  alt.  int.  A are  equal). 

But  ZAFG  — A FGD.  Hyp. 

Hence  AEFG=/.AFG.  Ax.  l. 

/.  KL  coincides  with  AH. 

But  KL  ||  CD.  Constr 

Hence  AB  ||  CD, 

(for  AB  coincides  with  EL,  which  is  ||  CD). 

Q.  E.  Eo 

Ex.  1 . If  Z BFG  = Z FGC,  prove  that  AB  and  CD  are  parallel. 

Ex.  2.  By  exact 
use  of  ruler  and 
eompasses,  at  a 
given  point  (P)  in  jq. 
a given  straight  line 
( OA ) construct  an  angle  equal  to  a given  A(B). 

Ex.  3.  By  exact  methods,  through  a given  point  draw  a line  paral- 
lel to  a given  line. 


60 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXVI.  Theorem 

126.  If  two  parallel  lines  are  cut  by  a transversal , the 
exterior  interior  angles  are  equal. 


Given  the  ||  lines  AB  and  CD  cut  by  the  transversal  PQ 


at  the  points  F and  G respectively. 

To  prove 

Z PFB  Z FGD. 

Proof. 

ZPFB  = ZAFG. 

(Why? ) 

ZFGD  = Z AFG, 

(being  alt.  int.  A of  parallel  lines) . 

Art.  124. 

Z PFB  = Z FGD. 

Ax.  1. 

In  like  manner  it  may  be  shown  that  ZPFA  = Z FGC. 

Q.  E.  D. 


Prop.  XXVII.  Theorem  (Converse  of  Prop.  XXVI) 

127.  If  two  straight  lines  are  cut  by  a transversal,  mah- 
ing  the  exterior  interior  angles  equal,  the  two  straight  lines 
are  parallel. 

Given,  on  Fig.  of  Prop.  XXV,  ZPFB  = ZFGD. 

To  prove  AB  ||  CD. 

Let  the  pupil  supply  the  proof. 

Ex.  If,  in  the  Fig.  to  Prop.  XXVT,  Z PFB  equals  67°,  find  the  otht* 
jeven  angles  in  the  figure  without  measuring  them. 


PABALLEL  LINES 


61 


Proposition  XXVIII.  Theorem 

128.  If  two  parallel  lines  are  cut  by  a transversal , the 
i turn  of  the  interior  angles  on  the  same  side  of  the  trans- 
versal is  equal  to  two  right  angles. 


Given  the  ||  lines  AB  and  CD  cut  by  the  transversal  PQ 


at  the  points  F and  G respectively. 

To  prove  A BFG  A AFGD  = 2 rt.  A. 

Proof.  Z FGD  = Z PFB.  Art.  126. 

To  each  of  these  equals  add  Z BFG. 

Then  Z BFG  A Z FGD=  Z PFB  + Z BFG.  Ax.  2. 
But  A PFB  + Z BFG  = 2 rt.  A.  Art.  73. 

.*.  Z BFG  A Z FGD  — 2 rt.  Z.  Ax.  1. 


Q.  E.  D. 

Prop.  XXIX.  Theorem  (Converse  of  Prop.  XXVIII ) 

129.  If  two  straight  lines  are  cut  by  a transversal , 
making  the  sum  of  the  interior  angles  on  the  same  side  of 
the  transversal  equal  to  two  right  angles , the  tiro  lines  are 
varallel. 

Given,  on  Fig.  of  Prop.  XXV,  ABFG  A Z FGD  = 2 rt.  4 . 
To  prove  AB  ||  CD. 

Let  the  pupil  supply  the  proof. 

Ex.  If,  on  Fig.  of  Prop,  XXVIII,  APFB  + AQGD= 180°,  are  AB 
nd  CD  |1  ? 


62 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXX.  Theorem 


130.  Two  angles  whose  sides  are  parallel,  each  to  each 
are  either  equal  or  supplementary. 


Given  AB  ||  DH,  and  BC  1 1 GF. 

To  prove  A ABC,  DBF  and  GEE  equal,  and  AAB6 
and  DEG  supplementary. 

Proof.  Produce  the  lines  BC  and  HD  to  intersect  in  P. 
Then  A ABC  = Z QPR,  and  Z QPB  = ADEF,  Art.  126. 


(being  ext.  int.  A of  ||  lines) . 

:.  A ABC  = ADEF. 

(Why?) 

Also 

A DEF  = A GEH. 

(Why?) 

:.  A ABC  = AGED. 

(Why?) 

Again 

A DEF  and  DEG  are  supplementary. 

Art.  73. 

. A ABC  and  DEG  are  supplementary. 

Ax.  8. 

Q.  E.  D. 


131.  Note.  It  is  to  be  observed  that  in  the  above  theorem  the 
two  angles  are  equal  if,  in  the  pairs  of  parallel  sides,  both  pairs  extend 
in  the  same  direction  from  the  vertices  ( A B and  DBF),  or  both  pairs  in 
opposite  directions  (AB  and  GER ) ; and  that  they  are  supplementary  if 
one  pair  extends  in  the  same  direction,  and  the  other  pair  in  opposite 
directions  ( A B and  DEG) . The  directions  of  the  sides  are  deter- 
mined by  connecting  the  vertices  of  the  angles  and  observing  whether 
the  lines  considered  lie  on  the  same  side  or  on  opposite  sides  of  the 
line  drawn. 


PARALLEL  LINES 


63 


Proposition  XXXI.  Theorem 

132.  Two  angles  whose  sides  are  perpendicular  each  to 
each  are  either  equal  or  supplementary . 


Given  BA  X EF,  and  BG  X BB'. 

To  prove  Z ABC  = Z FEB,  and  A ABO  and  FEB’  sup- 
plementary. 

Proof.  At  F let  line  FK  be  drawn  X FF  and  in  the 
same  direction  with  BA\  also  EG  X BB'  and  in  the  same 


direction  with  BG. 

Then  BA  1 1 FK,  and  BG  ||  EG,  Art.  121. 

( tivo  straight  lines  in  the  same  plane,  X the  same  straight  line,  are  || ). 

.’.  Z ABC  = Z KEG,  Arts.  130,  131. 

( two  A whose  sides  are  |[,  each  to  each,  and  extend  in  the  same  direction 
from  the  vertices  are  = ). 

But  A KEG  is  complement  of  A BEK,  Art.  33. 

{for  A DEG  is  a rt.  Z by  constr.). 

Also  A FEB  is  complement  of  A BEK,  Art.  33. 

{for  AFEK  is  a rt.  Z by  constr.). 

:.  A FEB  = A KEG,  Art.  ?5. 

( complements  of  the  same  Z are  =). 

/.  AABC=  A FEB.  Ax.  1. 

But  A FEB'  is  supplement  of  A FEB.  Art.  34. 

.’.  A FEB'  is  supplement  of  A ABC.  Ax.  8. 

, Q.  E.  D. 


133.  Note.  In  the  above  theorem  the  two  angles  are  equal  if  the 
sides,  considered  as  rotating  about  the  vertices,  are  taken  in  the  same 
order  (thus  BC  is  to  the  right  of  BA,  and  ED  to  the  right  of  EF 
AABG  = AFED)\  but  the  angles  are  supplementary  if  the  corres- 
ponding sides  are  taken  in  the  opposite  order  (thus,  BC  is  to  the  right 
of  BA  but  ED'  is  to  the  left  of  EF Z A BC= supplement  of  Z FED’). 


64 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXXII.  Theorem 

134.  The  sum  of  the  angles  of  a triangle  is  equal  to  two 
right  angles. 


Given  the  A ABC. 

To  prove  Z A + Z B + Z BCA  = 2 rt.  Z . 

Proof.  Produce  the  side  AC  to  the  point  D and  through 
C let  the  line  CE  be  drawn  ||  AB. 


Then 

ZECD  + ZBCE  + ZBCA  = 2 rt.  Z, 

Art.  77. 

(the  sum  of  all  the  A about  a point  on  the  same  side  of  a straight  line 
passing  through  the  point  = 2 rt.  A). 

But 

ZECB  = ZA, 

( being  ext.  mi.  A of  parallel  lines). 

Art.  126. 

Also 

ZBCE  = ZB, 

( being  alt.  int.  A of  parallel  lines). 

Art.  124. 

Substituting  for  ZECB  its  equal,  Zl,  and  for  Z BCE 
its  equal,  Z B,  Ax- 


Z1+  ZB  + ZBCA  = 2 rt.  Z. 

o.  E.  D. 

135.  Cor.  1.  An  exterior  angle  of  a triangle  is  equal  to 
the  sum  of  the  two  opposite  interior  angles. 

136.  Cor.  2.  The  sum  of  any  two  angles  of  a triangle 
is  less  than  two  right  angles. 

137.  Cor.  3.  In  a right  triangle  the  sum  of  the  two 
acute  angles  equals  one  right  angle . 


PARALLEL  LINES 


65 


138.  Cor.  4.  A triangle  can  have  hut  one  right , or  one 
obtuse  angle. 

139.  Cor.  5.  If  tivo  angles  of  one  triangle  equal  two 
angles  of  another  triangle , the  third  angle  of  the  first  tri 
angle  equals  the  third  angle  of  the  second. 

140.  Cor.  6.  If  an  acute  angle  of  one  right  triangle 
equals  an  acute  angle  of  another  right  triangle,  the  remain- 
ing acute  angles  of  the  triangles  are  equal. 

141.  Cor.  7.  Two  triangles  are  equal  if  two  angles  and 
a side  of  one  are  equal  to  two  angles  and  the  homologous  side 
of  the  second. 

142.  Cor.  8.  Two  right  triangles  are  equal  if  a leg  and 
an  acute  angle  of  one  are  equal  to  a leg  and  the  homologous 
acute  angle  of  the  other. 


Ex.  1.  If  two  angles  of  a triangle  are  56°  and  62°,  find  the  re- 
maining angle. 

Ex.  2.  If  one  acute  angle  of  a right  triangle  is  36°  15r,  find  the 
other  acute  angle. 

Ex.  3.  How  many  degrees  in  each  angle  of  an  equilateral  triangle? 

Ex.  4.  How  many  degrees  in  each  acute  angle  of  an  isosceles 
right  triangle? 

Ex.  5.  Is  it  possible  to  have  a triangle  whose  angles  are  45° 
62°,  72° ? 

Ex.  6.  If  one  angle  of  a triangle  is  42°,  find  the  sum  of  the  other 
two  angles. 

Ex.  7.  If  two  angles  of  a triangle  are  38°  and  65°,  find  all  the 
exterior  angles  of  the  triangle. 

Ex.  8.  If  the  vertex  angle  of  an  isosceles  triangle  is  38°,  find  each 
angle  at  the  base. 

Ex.  9.  If  an  angle  at  the  base  of  an  isosceles  triangle  is  50°,  find 
the  vertex  angle. 

Ex.  10.  An  exterior  angle  at  the  base  of  an  isosceles  triangle  is 
102°,  find  all  the  angles  of  the  triangle. 


E 


66 


BOOK  I,  PLANE  GEOMETRY 


QUADRILATERALS 

143.  A quadrilateral  is  a portion  of  a plane  bounded 
oy  four  straight  lines. 

The  sides  of  a quadrilateral  are  the  bounding  lines;  the 
angles  are  the  angles  made  by  the  bounding  lines;  the 
vertices  are  the  vertices  of  the  angles  of  the  quadrilateral. 

The  perimeter  of  a quadrilateral  is  the  sum  of  the  sides. 

144.  A diagonal  is  a straight  line  joining  two  vertices 
that  are  not  adjacent. 

145.  A trapezium  is  a quadrilateral  no  two  of  whose 
sides  are  parallel. 

146.  A trapezoid  is  a quadrilateral  which  has  two,  and 
only  two,  of  its  sides  parallel. 

147.  A parallelogram  is  a quadrilateral  whose  opposite 
sides  are  parallel. 


Trapezium  Trapezoid  Parallelogram 


148.  A rhomboid  is  a parallelogram  whose  angles  are 
oblique  angles. 

149.  A rhombus  is  a rhomboid  whose  sides  are  equal. 

150.  A rectangle  is  a parallelogram  whose  angles  are 
right  angles. 


151.  A square  is  a rectangle  whose  sides  are  equal. 


.Rhomboid  Rhombus  Rectangle  Square 


QUADRILATERALS 


67 


152.  The  base  of  a parallelo- 
gram is  the  side  upon  which  it  is 
supposed  to  stand,  as  AB.  The 
opposite  side  is  called  the  upper 
base  [CD). 

The  altitude  of  a parallelogram  is  the  perpendicular  dis- 
tance between  the  bases,  as  EF. 

153.  The  bases  of  a trapezoid  are  its  two  parallel  sides. 
The  legs  of  a trapezoid  are  the  sides  which  are  not  parallel. 
The  altitude  of  a trapezoid  is  the  perpendicular  distance  be- 
tween the  bases.  The  median  of  a trapezoid  is  the  line  join- 
ing the  midpoints  of  the  legs. 

154.  An  isosceles  trapezoid  is  a trapezoid  whose  legs 
are  equal. 


Ex.  1.  Draw  a quadrilateral  with  three  acute  angles  and  one  ob- 
tuse angle. 

Ex.  2.  Is  every  rhombus  a rhomboid  ? Is  every  rhomboid  a 
rhombus  ? 


Ex.  3.  What  is  the  difference  between  a square  and  a rhombus? 
What  properties  do  they  have  in  common? 

Ex.  4.  Find  the  perimeter  of  a square  foot  in  inches. 


By  aid  of  the  following  classification : 


Quadrilateral  . 


Trapezium. 
Trapezoid  . . 

„ Parallelogram 


Isosceles  trapezoid. 
Rectangle  . . . square. 
Rhomboid  . . . rhombus 


Ex.  5.  Determine  what  four  names  the  rhombus  is  entitled  to. 

Ex.  6.  Determine  what  properties  the  rhombus,  square  and 
rectangle  have  in  common. 

Ex.  7.  A diagonal  of  a rhombus  divides  the  rhombus  into  how 
many  triangles  ? What  kind  of  triangles  are  these  ? 


68 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXXIII.  Theorem 

io5.  The  opposite  sides  of  a parallelogram  are  equal,  and 
its  opposite  angles  are  also  equal. 


Given  the  parallelogram  ABCD. 

To  prove  AD  = BC,  AB  — DC,  /B=  /D,  and  /BAD  = 
/.BCD. 

Proof.  Draw  the  diagonal  AC. 

Then,  in  the  A ABC  and  ADC , 

AC— AC,  (Why?; 

Z BCA  = / CAD,  Art.  124. 

( being  alt.  int.  A of  parallel  lines). 

/ BA C = fACD, 

(■same  reason). 

:.  A ABC  = A ACD,  Art.  97. 

( two  A are  equal  if  two  A and  the  included  side  of  one  are  equal 
respectively  to  two  A and  the  included  side  of  the  other). 

:.  AD  = BC,  AB=DC  and  fB=  /D, 

( homologous  parts  of  equal  A ) . 

In  like  manner,  by  drawing  the  diagonal  BD,  it  may  be 
proved  that  /BAD=  / BCD.  g e d 

156.  Cor.  1.  A diagonal  divides  a parallelogram  into 
two  equal  triangles. 

157.  Cor.  2.  Parallel  lines  comprehended  between  par- 
allel lines  are  equal. 

158.  Cor.  3.  Two  parallel  lines  are  everywhere  equi- 
distant.   

Ex.  1.  In  the  above  figure,  prove  A BAD—  ABCD  by  use  of  Ajs.  2. 
Ex.  2.  Prove  the  opposite  angles  of  a parallelogram  equal,  Dy  use 
of  Art.  130. 


QUADRILATERALS 


69 


Proposition  XXXIV.  Theorem 

1 59.  If  the  opposite  sides  of  a quadrilateral  are  equal , 
the  figure  is  a parallelogram. 


Given  the  quadrilateral  ABCB  in  which  AB  — CB  and 
BC=AB. 

To  prove  ABCD  a L — / . 


Proof.  Draw  the  diagonal  AC. 
Then,  in  the  A ABC  and  ABC, 


AC=AC. 

(Why  ?) 

BC=AD. 

(Why  ?) 

AB=CB. 

(Why?) 

A ABC—  A ABC. 

(Why?) 

Z BAC  = Z ACB. 

(Why?) 

:.  AB  ||  CB, 

\if  two  lines  are  cut  by  a transversal,  making  the  alt.  int. 

lines  are  || ) . 

Art.  125. 
A equal,  the 

Also  Z BCA  =!  Z CAB. 

(Why  ?) 

:.  BC II  AB. 

Art.  125. 

:.  ABCB  is  a A7  , 

(a  CO  is  a quadrilateral  ivhose  opposite  sides  are 

Art.  147, 

II). 

Q.  E.  D, 


4 


70  BOOK  I.  PLANE  GEOMETRY 

Proposition  XXXV.  Theorem 

160.  If  two  sides  of  a quadrilateral  are  equal  and  par- 
allel, the  other  two  sides  are  equal  and  parallel  and  the  figure 
is  a parallelogram. 


Given  the  quadrilateral  ABCD  in  which.  BC  = and  I!  ID, 
To  prove  ABCD  a L 17  . 

Proof.  Draw  the  diagonal  AC. 

Then,  in  the  A ABC  and  ADC , 

AC=AC. 

BC=AD. 

ZBCA  = Z CAD, 

( being  alt.  int.  A of  parallel  lines), 

A ABC=A  ADC. 

:.  Z BAC=  Z.ACD. 

:.  AB  ||  CD, 

'■{if  two  lines  are  cut  by  a transversal,  making  the  alt.  int.  A equal,  the 

lines  are  || ) . 

.*.  ABCD  is  a CD  . Art.  147. 

Q.  E.  D. 

Ex.  1.  Show  that  in  a ZZ7  each  pair  of  adjacent  angles  is  sup- 
plementary. 

Ex.  2.  One  angle  of  a parallelogram  is  43°;  find  the  other  angles. 

Ex.  3.  If,  in  the  triangle  ABC,  ZM-60°,  AB  = 70°,  which  is  the 
longest  side  in  the  triangle  ? Which  the  shortest  ? 


(Why  ?) 
(Why  ?) 
Art.  124. 

(Why?) 
(Why  ?) 
Art.  125. 


QUADRILATERALS 


71 


Proposition  XXXVI.  Theorem 
161.  The  diagonals  of  a parallelogram  bisect  each  other . 


Given  the  diagonals  AC  and  BD  of  the  ZZ7  ABCD, 
intersecting  at  F. 

To  prove  AF=FC,  and  BF=FD. 

Proof.  Let  the  pupil  supply  the  proof. 

[Sug.  In  the  A BFC  and  AFD  what  sides  are  equal,  and  why  ? 
What  A are  equal,  and  why?  etc.] 

Q.  E.  D. 


Ex.  1.  How  many  pairs  of  equal  triangles  are  there  in  the  above 
figure  ? 

Ex.  2.  If  one  angle  of  a parallelogram  is  three  times  another  angle, 
find  all  the  angles  of  the  parallelogram. 

Ex.  3.  If  two  angles  of  a triangle  are  p°  and  q°,  find  the  third 
angle. 

Ex.  4.  If  two  angles  of  a triangle  are  x°  and  90°  + x°,  find  the 
third  angle. 

Ex.  5.  If  one  angle  of  a parallelogram  is  a°,  find  the  other  angles. 

Ex.  6.  Construct  exactly  an  angle  of  60°. 

Ex.  7.  How  large  may  the  double  of  an  acute  angle  be  ? how 
small  ? 

Ex.  8.  How  large  may  the  double  of  an  obtuse  angle  be  ? how 
small  ? 


72 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXX YII.  Theorem 

162.  Two  parallelograms  are  equal  if  two  adjacent  sides 
and  the  included  angle  of  one  are  equal,  respectively , to  two 
adjacent  sides  and  the  included  angle  of  the  other. 


Given  the  CO  ABCD  and  A'B'C'D'  in  which  AB=A'B', 
AD=A'D',  and  Z A = Zl'. 

To  prove  CO  ABCD  = CO  A'B'C'D'. 

Proof.  Apply  the  CO  A'B'C'D'  to  the  £17  ABCD  so  that 
A'D'  shall  coincide  with  its  equal  AD. 

Then  A'B'  will  take  the  direction  of  AB  ( for  /i'=ZA); 
and  point  B'  will  fall  on  B ( for  A'B'=AB). 

Then  B'C  and  BC  will  both  be  ||  AD  and.  will  both 
pass  through  the  point  B. 

B'C'  will  take  the  direction  of  BC,  Geom.  Ax.  3. 
( through  a given  point  one  straight  line,  and  only  one,  can  he  drawn  j| 
another  given  straight  line). 

In  like  manner,  D'C'  must  take  the  direction  of  DC. 

C must  fall  on  C,  ' Art.  64 

( two  straight  lines  can  intersect  in  but  one  point). 

CO  ABCD  = CO  A'B'C'D',  Art.  47. 

{geometric  figures  which  coincide  are  equal). 

Q.  E.  D. 

163.  Cor.  Two  rectangles  which  have  equal  bases  and 
equal  altitudes  are  equal. 


Ex.  Construct  exactly  an  angle  of  30°. 


POLYGONS 


73 


POLYGONS 


164.  A polygon  is  a portion  of  a plane  bounded  by 
straight  lines,  as  ABODE. 

The  sides  of  a polygon  are  its  bounding  lines;  the 
perimeter  of  a polygon  is  the  sum  of  its  sides;  the  angles 
of  a polygon  are  the  angles  formed  by  its  sides;  the 
vertices  of  a polygon  are  the  vertices  of  its  angles. 

A diagonal  of  a polygon  is  a straight  line  joining  two 
vertices  which  are  not  adjacent,  as  BD  in  Fig.  1. 


165.  An  equilateral  polygon  is  a polygon  all  of  whosg 
sides  are  equal. 

166.  An  equiangular  polygon  is  a polygon  all  of  whose 
angles  are  equal. 

What  four- sided  polygon  is  equilateral  but  not  equi- 
angular ? Also,  wdiat  four-sided  polygon  is  both  equilateral 
and  equiangular? 

167.  A convex  polygon  is  a polygon  in  which  no  side, 
if  produced,  will  enter  the  polygon,  as  ABODE  (Fig.  1). 

Each  angle  of  a convex  polygon  is  less  than  two  right 
’ angles  and  is  called  a salient  angle. 

168.  A concave  polygon  is  a polygon  in  which  two  or 
more  sides,  if  produced,  will  enter  the  polygon,  as  FGEIJK 


B 


A. 


E 

Fig.  1 


K 

Fig.  2 


(Fig.  2). 


74 


BOOK  I.  PLANE  GEOMETRY 


Some  angle  of  a concave  polygon  must  be  greater  than 
two  right  angles,  as  angle  GHI  of  Fig.  2.  Such  an  angle 
is  termed  a re-entrant  angle. 

If  the  kind  of  polygon  is  not  specified  in  this  respect,  a 
convex  polygon  is  meant. 

169.  Two  mutually  equiangular  polygons  are  polygon* 

whose  corresponding  angles  are  equal,  as  Figs.  3 and  4. 


170.  Two  mutually  equilateral  polygons  are  polygons 

whose  corresponding  sides  are  equal,  as  Figs.  5 and  6. 

From  Figs.  3 and  4 it  is  seen  that  two  polygons  may  be 
mutually  equiangular  without  being  mutually  equilateral. 

What  similar  truth  maybe  inferred  from  Figs.  5 and  6'? 

171.  Names  of  particular  polygons.  Some  polygons  are 
used  so  frequently  that  special  names  have  been  given  to 
them . A polygon  of  three  sides  is  called  a triangle ; one  of 
four  sides,  a quadrilateral ; one  of  five  sides,  a pentagon ; of 
six  sides,  a hexagon ; of  seven  sides,  a heptagon ; of  eight 
sides,  an  octagon  ; of  ten  sides,  a decagon  ; of  twelve  sides, 
a dodecagon;  of  fifteen  sides,  a pentedecagon ; of  n sides,  an 
n-gon. 

Ex.  1.  Let  the  pupil  illustrate  Arts.  169  and  170  by  drawing  two 
pentagons  that  are  mutually  equilateral  without  being  mutually  equL 
angular,  and  another  pair  of  which  the  reverse  is  true. 

Ex.  2.  Can  two  triangles  be  mutually  equilateral  without  being 
mutually  eauiangular  ? What  polygons  can  ? 

Ex.  3.  How  does  the  number  of  vertices  in  a polygon  compare 
with  the  number  of  sides  f 


POLYGONS 


75 


Proposition  XXXVIII.  Theorem 

172.  The  sum  of  the  angles  of  any  polygon  is  equal  to 
two  right  angles  taken  as  many  times,  less  tivo,  as  the  poly- 
gon has  sides. 


3 triangles  4 triangles  5 triangles 


Given  a polygon  of  n sides  (the  above  polygons  of  5, 
6,  7 sides  being  used  merely  as  particular  illustrations,  to 
aid  in  carrying  forward  the  proof) . 

To  prove  the  sum  of  its  A =(n — 2)  2 rt.  A. 

Proof.  By  drawing  diagonals  from  one  of  its  vertices 
the  polygon  is  divided  into  ( n — 2)  triangles. 

Then  the  sum  of  the  A of  each  triangle  = 2 rt.  A . Art.  134. 
( the  sum  of  the  A of  a A is  equal  to  2 rt.  A). 

Hence  the  sum  of  the  A of  the  (n — 2)  A = (w  — 2)  2 rt.  A . 

Ax.  4. 

But  the  sum  of  the  A of  the  polygon  is  equal  to  the 
sum  of  the  A of  the  ( n — 2)  A.  Ax.  8. 

Hence  the  sum  of  the  A of  the  polygon  = in  — 2)  2 rt.  A . 

Ax.  I. 

Q.  E.  D. 

173.  Cor.  1.  The  sum  of  the  angles  of  a polygon  equals 
(2 » — 4)  rt.  A . 

174.  Cor.  2.  In  an  equiangular  polygon  of  n sides 

, 7 7 (n — 2)  2 rt.  A 2 n — 4 , . 

each  angle  equals  , or rt.  A . 

n n 


76 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XXXIX.  Theorem 

175.  The  sum  of  the  exterior  angles  of  a polygon  formed 
by  producing  its  sides  in  succession  at  one  extremity  equods 
four  right  angles. 


Given  a polygon  of  n sides  having  its  sides  produced 
in  succession. 

To  prove  the  sum  of  the  exterior  A =4  rt.  A . 

Proof.  If  the  interior  A of  the  polygon  be  denoted  by 
A,  B,  C,  and  the  corresponding  exterior  A by  a,  b,  c, 

A A + Aa= 2 rt.  A,  (Why?) 

AB  + Ab  = 2rt.  A,  (Why?) 

etc. 

Adding,  int.  A + ext.  A = n times  2rt.  A = 2«rt.  A . Ax.  2 
But  int.  A = (n — 2)  2 rt.  A =2  n rt.  A — 4rt.  A . Art.  173 
Ext.  A =4  rt.  A . 

q.  e.  r 


Ex.  1.  What  does  the  sum  of  the  interior  angles  of  a hexagoi 
equal  ? of  a heptagon  ? of  a decagon  ? 

Ex.  2.  Each  angle  of  an  equiangular  pentagon  contains  how  many 
degrees  ? of  an  equiangular  hexagon  ? octagon  ? decagon  ? 

Ex.  3.  Would  a quadrilateral  constructed  of  rods  binged  at  the 
ends  (i.  e.,  at  the  vertices  of  the  quadrilateral)  be  rigid?  Would  a 
triangle  so  constructed  be  rigid  ? Would  a pentagon  ? 


MISCELLANEOUS  THEOREMS 


7? 


MISCELLANEOUS  THEOREMS 

Proposition  XL.  Theorem 

176.  If  three  or  more  parallels  intercept  equal  parts  on 
one  transversal,  they  intercept  equal  parts  on  every  trans 
ve^sol. 


Given  AP,  BQ , CR  and  DT  parallel  lines  intercepting1 
equal  parts  AB , BC  and  CD  on  the  transversal  AD. 

To  prove  that  they  intercept  equal  parts  PQ,  QR  and 
RT  on  the  transversal  PT. 

Proof.  Through  A,  B and  C let  A Mi  BG  and  CR  be 
drawn  parallel  to  PT  and  meeting  the  lines  BQ.  CR  and  DT 
in  the  points  F,  G and  R respectively. 

Then  the  lines  AF,  BG  and  CR  are  l|  , Art.  122. 

( two  straight  lines  ||  a third  straight  line  are  ||  each  other). 

.In  the  A ABF,  BCG  and  CDR,  Z ABF  = Z BCG  = 

/ CDR,  , „ Art.  126 

( being  ext.  int.  A of  ||  lines). 

Also  Z.BAF—  ZCBG=  Z.DCR,  [same  reason) . 


And  AB  — BC=CD.  Hyp. 

/.  A ABF=  A BCG=A  CDR.  Art.  97. 

AF=BG=CR.  (Why?) 

But  AF=PQ,  BG=QR,  CR  = RT,  Art.  157. 

{parallel  lines  comprehended  between  parallel  lines  are  equal). 

PQ=QR=RT.  ax.  l. 

0,  E.  D. 


78 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XLI.  Theorem 

177.  The  line  which  joins  the  midpoints  of  two  sides  of 
a triangle  is  parallel  to  the  third  side,  and  is  equal  to  one- 
half  the  third  side. 


A 


Given  D the  midpoint  of  AB,  and  E the  midpoint  of 
AC  in  the  triangle  ABC. 

To  prove  IDE  ]|  BC  and  =$BC. 

Proof.  Through  B let  BL  be  drawn  H AC,  and  meeting 


BE  produced  at  L. 

Then,  in  the  A ABE  and  BDL, 

AB  — BB,  (Why?) 

Z ABE  = Z BBL.  (Why  T) 

ZBAE=  ZBBL.  (Why?) 

/.  AABE±  ABBL.  (Why?) 

.*.  BE—BL,  or  BE—hLE,  and  AE=BL.  (Why?) 
But  AE=EC  (Hyp.)  .*.  EC=BL.  As.  1. 

Also  EC  ||  BL.  Constr. 

BLEC  is  a ZZ7  , Art.  160. 

{if  two  sides  of  a quadrilateral  are  equal  and  parallel  the  figure  is  a / 7 ). 

.-.  BE  ||  BC.  Art.  147. 

Also  LE  — BC,  {opp.  sides  of  a ED  are  =).  Art.  155. 

/.  J LE,  or  BE=h  BC.  As.  5. 

Q.  E.  D. 


178.  Cor.  The  line  ivhich  bisects  one  side  of  a triangle 
and  is  parallel  to  another  side  bisects  the  third  side. 

Thus,  given  AB  = BB  and  BE  ||  BC,  then  will  AE=EC. 
For,  suppose  a line  FO  drawn  through  A ||  BC, 


MISCELLANEOUS  THEOREMS 


79. 


Then  the  three  parallels  FG,  BE,  BC  will  intercept 
equal  parts  on  AB.  Hyp. 

they  intercept  equal  parts  on  AG.  Art.  176. 
AE=EC. 


Proposition  XLII.  Theorem 

2 79.  The  line  which  joins  the  midpoints  of  the  legs  of 
a trapezoid  is  parallel  to  the  bases  and  equal  to  one-half 

iheir  sum. 


Given  the  trapezoid  ABCB,  E the  midpoint  of  the  leg 
AB,  and  F the  midpoint  of  the  leg  CD. 

To  prove  that  a line  joining  E and  F is  [|  AD  and  BG 
and  = i ( AD  + BC) . 

Proof.  Draw  the  diagonal  BD  and  take  G its  mid- 
point. Draw  EG  and  GF. 

Then,  in  the  A ABD,  EG  II  AD  and  = J AD,  Art.  177. 
'.lie  line  which  joins  the  midpoints  of  two  sides  of  a A is  ||  the  third  side 
and  = one-half  the  third  side). 

Also,  in  the  A BDG,  GF  I!  BG  and  = i BC,  ( same  reason). 

GF  and  AD  both  j|  BC;  GF  ||  AD,  Art.  122. 

(two  straight  lines  ||  a third  straight  are  ||  each  other). 

. •.  EG  and  GF  are  both  ||  AD. 

. \ EG  and  GF  form  one  and  the  same  straight  line  EF, 

Geom.  Ax.  3, 

.through  a given  point  one  line,  and  only  one,  can  he  drawn  ||  to  another 

given  line) . 

:.  EF  ||  AD  and  BG. 

Also  EG  = v AD,  and  GF  = £ BC. 

Adding,  EG  + GF,  or  EF  = i (AD  + BC) . Ax.  2. 

Q.  E.  D. 

180.  Cor.  A line  drawn  bisecting  one  leg  of  a trapezoid 
and  parallel  to  the  base  bisects  the  other  lea  also. 


80  BOOK  I.  PLANE  GEOMETRY 


Proposition  XLIII.  Theorem 

181.  The  bisectors  of  the  angles  of  a triangle  intersect  at 
a common  point  ( called  the  in-center). 


Given  the  A ABC  with  the  lines  AP,  BQ,  CR  (Fig-  1) 
bisecting  the  Z«  BAC,  ABC,  ACB,  respectively. 

To  prove  that  AP,  BQ , CR  intersect  in  a common  point. 

Proof.  Let  AP,  the  bisector  of  / BAC,  and  CR,  the 
bisector  of  Z BCA , intersect  at  the  point  0. 

Then  0,  being  in  bisector  AP,  is  equidistant  from  AB 
and  AC,  Art.  117. 

(i every  point  in  the  bisector  of  an  Z is  equidistant  from  the  sides  of  the  Z ). 

Also  0,  being  in  bisector  CR,  is  equidistant  from  AC  and 
BC,  {same  reason) . 

Hence  0 is  equidistant  from  the  sides  AB  and  BC.  Ax.  1. 

.-.  0 is  in  the  bisector  of  /.ABC,  Art.  117. 

{every  point  equidistant  from  the  sides  of  an  Z lies  in  the  bisector  of  the  Z ) . 

Hence  BQ,  or  BQ  produced,  passes  through  0. 

Hence  the  bisectors,  AP,  BQ,  CR,  of  the  three  / of  the 
A ABC  intersect  at  0.  Q.  E.  D. 

182.  Cor.  The  point  in  which  the  three  bisectors  of  the 
angles  of  a triangle  iritersect  is  equidistant  from  the  three  sides 
of  the  triangle. 

183.  Def.  Concurrent  lines  are  lines  which  pass  through, 

the  same  point.  


Ex.  Find  other  Z of  above  figure  if  / BAC  = 72°  and  Z BCA  = 44°. 


MISCELLANEOUS  THEOREMS 


81 


Proposition  XLIY.  Theorem 

1 84.  The  perpendicular'  M sectors  of  the  sides  of  a trian- 
gle intersect  at  a common  point/ {called  the  circumcenter). 


Given  the  A ABC  with  DP,  EQ,  FR  (Fig.  1),  the  X 
bisectors  of  the  sides  AB,  BC,  CA,  respectively. 

To  prove  that  DP,  EQ,  FR  intersect  at  a common  point. 
Proof.  Let  DP  and  EQ  intersect  at  the  point  0. 

Then  0,  being  in  X DP,  is  equidistant  from  the  points  A 

and  B,  Art.  112. 

{every  point  in  the  perpendicular  bisector  of  a line  is  equally  distant  from 
the  extremities  of  the  line ) . 

Also  0,  being  in  X EQ,  is  equidistant  from  the  points 
B and  C, 

{same  reason ). 

Hence  0 is  equidistant  from  A and  C.  Ax.  1. 

.*.  0 is  in  the  X bisector  of  AC,  Art.  115. 

{the  X bisector  of  a line  is  the  locus  of  all  points  equidistant  from  the 
extremities  of  the  line). 

Hence  FR,  or  FR  produced,  passes  through  0. 

Hence  the  perpendicular  bisectors,  DP,  EQ,  FR,  of  the 
three  sides  of  the  A ABC  meet  in  the  point  0.  q.  e.  d. 

185.  Cor.  The  point  in  which  t,he  perpendicular  bisectors 
of  the  sides  of  a triangle  meet  is  equidistant  from  the  vertices 
of  the  triangle. 

F 


82 


BOOK  I.  PLANE  GEOMETRY 


Proposition  XLV.  Theorem 

186.  The  perpendiculars  from  the  vertices  of  a triangle 
to  the  opposite  sides  meet  in  a point  ( called  the  ortho-center). 


Given  AD,  BF,  and  CE  the  perpendiculars  from  the  ver- 
tices A,  B , and  C of  the  A ABC  to  the  opposite  sides. 

To  prove  that  AD,  BF,  and  CE  intersect  in  a common 
point. 

Proof.  Through  A,  B,  and  C let  the  lines  PR,  PQ  and 
QR  be  drawn  ||  BC,  AC,  and  AB,  respectively,  and  forming 
the  A PQR. 

Then  AD  A PR,  Art,  123. 

( for  AD  X BC,  and  a line  X one  of  two  ||  lines  is  X the  other  also) . 

Also  APBC  and  ABCR  are  EU  . Constr. 

.-.  AP  = BC,  and  AR  = BC,  Art.  155. 

( the  opposite  sides  of  a CD  are  = ). 

.’.  AP  = AR.  Ax.  1. 

Hence,  in  the  A PQR,. AD  is  the  perpendicular  bisector 
of  side  PR. 

In  like  manner  it  may  be  shown  that  BF  is  the  perpen- 
dicular bisector  of  PQ,  and  that  CE  is  the  perpendicular 
bisector  of  QR. 

Hence  AD,  BF,  and  CE  are  the  perpendicular  bisectors 
of  the  sides  of  the  A PQR. 

AD,  BF,  and  CE  meet  in  a common  point,  Art.  184. 

(the perpendicular  bisectors  of  the  sides  of  a A are  concurrent) . 

Q.  E.  D. 


MISCELLANEOUS  THEOREMS 


S3 


Proposition  XL VI.  Theorem 

187.  The  medians  of  a triangle  intersect  ( or  are  con- 
current) in  a point  ( called  the  centroid)  loliich  cuts  off  two- 
thirds  of  each  median  from  its  vertex. 

~B 


d 

Given  AD,  BF  and  CF  the  three  medians  of  A ABC. 

To  prove  that  AD,  BF  and  CF  intersect  at  a point  which 
cuts  off  two- thirds  of  each  median  from  its  vertex. 

Proof.  Let  the  medians  AD  and  CF  intersect  at  0. 

Take  R the  midpoint  of  AO,  and  S the  midpoint  of  OC, 
and  draw  RF,  ED,  DS,  and  SR. 

Then  ED  \\  AC  and  — ^ AC.  Art.  177, 

Also,  in  the  A A OC, 

RS  ||  AC  and  = J AC.  Art.  177. 

Hence  ED  ||  RS  and  — RS  Art.  122  and  Ax.  l. 

.-.  REDS  is  a EJ  . (Why  ?) 

Hence  FS  and  RD  bisect  each  other.  (Why?) 

But  AR  — RO,  and  CS—SO.  Constr. 

.'.  AR  = RO=  OD,  and  CS  = S0=  OF.  Ax.  l. 

Hence  CF  crosses  AD  at  a point  b,  such  that  AO  = % AD. 
In  like  manner  it  may  be  shown  that  BF  crosses  AD  at 
the  point  0. 

Hence  the  medians  AD,  BF,  and  CE  intersect  at  point  0, 
which  cuts  off  two -thirds  of  each  median  from  its  vertex. 

Q.  £.  P, 


84 


BOOK  I.  PLANE  GEOMETEY 


188.  Properties  of  rectilinear  figures  to  be  proved  by  the 
pupil.  Proof  of  equality  of  triangles.  Other  properties  of 
rectilinear  figures  will  now  be  given  which  are  to  be  demon 
strated  by  the  pupil.  These  theorems  will  be  arranged  in 
groups,  according  to  the  method  of  proof  to  be  used 
followed  by  a group  of  general  or  mixed  exercises. 

Let  the  pupil  form  a list  of  the  conditions  that  mahe 
two  triangles  equal. 

(See  Arts.  96,  97,  98,  101,  102,  141,  142.) 


EXERCISES.  CROUP  4 

EQUALITY  OF  TBLANGLES 

Ex.  1.  Given  ABC  any  triangle,  BO  the  bi- 
sector of  A ABC,  and  AD  X BO ; prove  A ABO= 

A BOD. 

[Sug.  In  the  A ABO  and  DBG  -what  lines 
are  equal  f What  A are  equal  ? etc.] 

Ex.  2.  If,  at  any  point  in  the  bisector  of  an 
angle,  a L be  erected  and  produced  to  meet  the 
sides  of  the  angle,  how  many  triangles  are  formed  ? 

Are  these  triangles  equal  ? Prove  this. 

Ex.  3.  If,  through  the  midpoint  of  a given 
straight  line,  another  line  be  drawn,  and  produced 
to  meet  the  perpendiculars  erected  at  the  ends 
of  the  given  line,  the  triangles  so  formed  are 
equal. 

' Ex.  4.  If  two  straight  lines  bisect  each  other  and  their  extremi- 
' ties  be  joined,  how  many^  pairs  of  equal  triangles  are  formed? 
Prove  this. 

Ex.  5.  If  equal  segments  from  the  base  be  laid  off  on  the  sides 
of  an  isosceles  triangle,  and  lines  be  drawn  from  the  extremities  of  the 
segments  to  the  opposite  vertices,  prove  that  two  pairs  of  equal  tri- 
angles are  formed. 


EXERCISES.  EQUALITY  OF  TRIANGLES 


85 


Ex.  6.  If,  upon  the  sides  of  an  angle,  equal  segments  be  laid  off 
from  the  vertex,  and  lines  he  drawn  from  the  ends  of  these  segments 
to  any  point  in  the  bisector  of  the  angle,  prove  that  the  triangles 
formed  are  equal. 

Ex.  7.  If  two  sides  of  a triangle  be  produced,  each  its  own 
length,  through  the  vertex  in  which  they  meet,  and  the  extremities 
of  the  produced  parts  be  joined,  prove  that  a new  triangle  is  formed 
which  equals  the  original  triangle. 

Ex.  8.  Given  AB=DC,  and  BC=DA;  prove 
ABAC— ABAC.  What  other  pair  of  equal 
triangles  is  there  in  the  figure  ? 

Ex.  9.  Two  right  triangles  are  equal  if  their  corresponding  legs 
are  equal. 

Ex.  10.  The  altitudes  from  the  extremities  of  the  base  of  an 
isosceles  triangle  upon  the  legs  of  the  triangle  divide  the  figure  into 
how  many  pairs  of  equal  triangles  ? Prove  this. 

Ex.  11.  In  a given  quadrilateral  two  adjacent  sides  are  equal  and  a 
diagonal  bisects  the  angle  between  these  sides.  Prove  that  the  diagonal 
bisects  the  quadrilateral. 

Ex.  12.  If,  from  the  ends  of  the  shorter  base  of  an  isosceles  trape- 
zoid, lines  be  drawn  parallel  to  the  legs  and  produced  to  meet  the 
other  base,  prove  that  a pair  of  equal  triangles  is  formed. 


189.  Proof  of  the  equality  of  lines.  There  are  several 
methods  of  proving  that  two  lines  (segments)  are  equal. 
One  of  the  principal  methods  of  proving  that  two  lines  are 
equal  is  by  proving  that  two  triangles,  in  which  the  given 
lines  form  homologous  parts,  are  equal. 


86  BOOK  I.  PLANE  GEOMETRY 

EXERCISES.  CROUP  5 

EQUALITY  OF  LINES 

Ex.  1.  Given  ABC  any  triangle,  BO  the  bi- 
sector of  Z ABC,  and  AD  -L  BO;  prove  AO=OD. 

Ex.  2.  If,  at  any  point  in  the  bisector  of  an 
angle,  a perpendicular  be  erected  to  the  bisector 
and  produced  to  meet  the  sides  of  the  angle, 
the  perpendicular  is  divided  into  two  equal  parts  at  the  given  point 

Ex.  3.  If  two  sides  of  a triangle  be  produced, 
each  its  own  length,  through  the  common  vertex, 
the  line  joining  the  extremities  of  the  produced 
parts  equals  the  third  side  of  the  triangle  (i  e., 

BE  = BA). 

Ex.  4.  If  equal  segments  from  the  base  be  laid 
off  on  the  legs  of  an  isosceles  triangle,  lines  drawn  from  the  ends  of 
these  segments  to  the  opposite  vertices  are  equal. 

Ex.  5.  Given  AR  ||  QB,  and  AP=PB; 
prove  RP  = PQ. 

Ex.  6.  The  bisector  of  the  vertical  an- 
gle of  an  isosceles  triangle  bisects  the  base. 

Ex.  7.  The  altitudes  of  an  isosceles  triangle  upon  the  legs  are 
equal. 

Ex.  8.  The  diagonals  of  a rectangle  are  equal. 

Sx.  9.  The  medians  of  an  isosceles  triangle  to  the  legs  are  equal. 

Ex.  10.  If  two  altitudes  of  a triangle  are  equal,  the  triangle  is 
isosceles. 

Ex.  11.  The  perpendiculars  to  a diagonal 
of  a parallelogram  from  a pair  of  opposite 
vertices  are  equal. 

Ex.  12.  If  the  equal  sides  of  an  isosceles  triangle  be  produced 
through  the  vertex  so  that  the  produced  parts  are  equal,  the  lines  join- 
ing the  extremities  of  the  produced  parts  to  the  extremities  of  the 
base  are  equal. 

Ex.  13.  If  the  base  of  an  isosceles  triangle  be  trisected,  lines 
drawn  from  the  vertex  to  the  points  of  trisection  are  equal. 


A R 


Q 2 


EXERCISES.  EQUALITY  OF  ANGLES 


87 


Lines  may  also  be  proved  equal  by  showing  that  they  are: 
opposite  equal  angles  in  a triangle ; 
or  opposite  sides  of  a parallelogram ; 
or  para  llel  lines  comprehended  between  parallel  lines-  etc. 
(See  Arts.  100,  155,  157,  etc.). 


Ex.  14.  If  the  exterior  angles  at  the  base 
of  a triangle  are  equal,  the  trian- 
gle is  isosceles. 

Ex.  15.  In  the  bisectors  of 
the  equal  angles  of  an  isosceles 
triangle,  the  segments  next  to 
the  base  are  equal  (AO  = OP). 


D 


Ex  16.  In  A ABC,  given 
AB  = AC,  DE  ||  BC]  prove 
AD  = AE. 

Ex.  17.  Given  AB  = DC, 
and  BC= AD-,  prove  AE= EC. 


190.  Proof  of  the  equality  of  angles  may  be  obtained 
in  several  different  ways. 

One  of  the  principal  methods  of  proving  that  two  angles 
are  equal  is  by  proving  that  two  triangles,  in  which  the  given 
angles  form  homologous  parts,  are  equal. 


EXERCISES.  CROUP  6 


EQUALITY  OF  ANGLES 


Lx.  1.  Given  ABC  any  A,  BO  the  bisector 
of  the  A ABC,  and  AD  L BO]  prove  ABAO 
= Z BDO. 

Ex.  2.  If,  at  any  point  in  the  bisector  of  an 
angle,  a perpendicular  be  erected  and  produced 
to  me°t  the  sides  of  the  angle,  the  per- 
pendicular makes  equal  angles  with  the  sides 
of  the  angle. 

Ex.  3.  Given  AB  = DC,  and  AD  = BC\ 
prove  AB  = Z D. 


88 


BOOK  I.  PLANE  GEOMETRY 


Ex.  4.  If  equal  segments  from  the  base  be  laid  olf  on  the  legs  of 
an  isosceles  triangle,  the  lines  drawn  from  the  extremities  of  the 
segments  to  the  opposite  vertices  make  equal  angles  with  the  base. 

Ex.  5.  The  median  to  the  base  of  an  isosceles  triangle  is  perpen* 
dicular  to  the  base. 

Ex.  6.  The  altitudes  upon  the  legs  of  an  isosceles  triangle  make 
equal  angles  with  the  base. 

Ex.  7.  The  diagonals  of  a rhombus  bisect  its  angles. 


Angles  may  also  be  proved  equal  by  proving  tbat 
they: 

are  opposite  equal  sides  in  an  isosceles  triangle ; 
or  are  vertical  angles ; 

or  are  complements  {or  supplements ) of  equal  angles; 
or  bg  the  use  of  the  properties  of  parallel  lines; 
or  that  their  sides  are  parallel,  or  perpendicular . 

(See  Arts.  75,  78,  99,  124,  126,  130,  132.) 


Ex.  8.  In  an  isosceles  triangle  the  exterior  angles 
made  by  producing  the  base  are  equal. 


Ex.  9.  Given  AC  = CB,  and  BE  || 
AB ; prove  A CDE  = Z CEB. 

Ex.  10.  Given  CE\\  AB,  and  ADCE 
= Z EC  A ; prove  AB  = A A. 


Ex.  11.  Conversely  given 
A A — A B,  and  CE  ||  AB  ; 
prove  that  CE  bisects  A DC  A. 


Ex.  12.  Given  BD  the  bisector  of  the 
angle  ABC,  and  PR  H CB;  prove  PUB  an 
isosceles  A . Let  the  pupil  state  this  theorem 
in  general  language. 


EXERCISES.  PARALLEL  LINES 


89 


Ex.  13.  Given  AB  — AC, 
and BD—CE;  prove  /BCD— 
/ CBE.  How  maDy  pairs  of 
equal  A in  the  figure  ? of 
equal  lines?  of  equal  X ? 

Ex.  14.  A line  drawn 
through  the  vertex  of  an  angle, 
perpendicular  to  the  bisector 
of  the  angle,  makes  equal 
angles  with  the  sides  of  the  given  angle. 


Ex.  15.  If  a straight  line  which  bisects 
one  of  two  vertical  angles  be  produced,  it 
bisects  the  other  vertical  angle  also. 


191.  Proof  that  two  lines  are  parallel  may  be  obtained 
by  showing  that: 

the  lines  are  cut  by  a transversal , making  the  alter- 
nate interior  angles  equal ; 
or  making  the  exterior  interior  angles  equal; 
or  making  the  interior  angles  on  the  same  side  of  the 
transversal  supplementary ; 

or  that  the  lines  are  opposite  sides  of  a parallelogram ; 
or  that  one  of  the  lines  joins  the  midpoints  of  two  sides 
of  a triangle,  and  the  other  line  is  the  third  side  of 
the  triangle. 

(See  Arts.  125,  127,  129,  147,  177,  etc.) 


EXERCISES.  CROUP  7 

PARALLEL  LINES 

Ex.  1.  If  two  sides  of  a triangle  be  produced,  each  its  own  length, 
through  the  common  vertex,  the  line  joining  their  ex- 
tremities is  parallel  to  the  third 
side  of  the  triangle. 

Ex.  2.  The  bisectors  of  two 
alternate  interior  angles  of  par- 
allel lines  are  parallel. 


Ex.  3. 

prove  CE 


Given 
I BA. 


Z A = /LB,  and  Z BCE  = /.EC A ; 


90 


BOOK  I.  PLANE  GEOMETEY 


Ex.  4.  The  bisectors  of  the  opposite  angles  of  a parallelogram  are 
parallel. 

Ex.  5.  Lines  perpendicular  to  parallel  lines  are  parallel  (or 
coincide). 

Ex.  6.  Two  straight  lines  are  parallel  if  two  points  on  one  line 
are  equidistant  from  the  other  line. 

192.  The  proof  of  a numerical  property  of  the  recth. 
linear  figures  (of  Book  I)  usually  depends  on  one  of  the 
following: 

The  sum  of  the  angles  about  a given  point  on  the  sami 
side  of  a straight  line  passing  through  the  point  is  180° ; 

or  the  sum  of  the  angles  about  a point  is  360° ; 

or  the  sum  of  the  angles  of  a triangle  is  180°; 

or  the  sum  of  the  interior  angles  of  parallel  lines  on  the 
same  side  of  a transversal  is  180°; 

or  the  sum  of  the  interior  angles  of  a polygon  of  n sides 
is  (»  — 2)  180°; 

or  the  sum  of  the  exterior  angles  of  a polygon  is  360°. 

(See  Arts.  76,  77,  128,  134,  172,  175.) 

EXERCISES.  CROUP  8 

NUMERICAL  PROPERTIES 

Ex.  1.  If  an  exterior  angle  of  a triangle  is  123°  and  an  opposite 
interior  angle  is  38°,  find  the  other  two  angles  of  the  triangle. 

Ex.  2.  Find  the  angle  formed  by  the  bisectors  of  the  two  acute 
angles  of  a right  triangle. 

Ex.  3.  If  two  angles  of  a triangle  are  50°  and  60°,  find  the  angle 
formed  by  their  bisectors.  Find  the  same  if  the  two  angles  contain 
p°  and  q°. 

Ex.  4.  If  the  vertex  angle  of  an  isosceles  triangle  is  40°  and  a 
perpendicular  is  drawn  from  an  extremity  of  the  base  to  the  oppos:  o® 
side,  find  the  angles  of  the  figure. 


EXERCISES.  NUMERICAL  PROPERTIES 


91 


Ex.  5.  If  tlie  vertex  angle  of  an  isosceles  triangle  is  40°,  find  the 
angle  included  between  the  altitudes  drawn  from  the  extremities  of 
the  base  to  the  opposite  sides. 

Ex.  6.  How  many  degrees  in  each  angle  of  an  equiangular  dodeca- 
' gon?  of  an  equiangular  «-gon? 

Ex.  7.  How  many  diagonals  are  there  in  a pentagon?  in  a hexa- 
gon? in  a decagon?  inanw-gon? 

The  methods  of  proving  that  a given  angle  is  a right 
angle  (or  that  a given  line  is  perpendicular  to  another  given 
line ) , or  that  one  angle  is  the  supplement  of  another,  are 

closely  related  to  the  above  methods  of  obtaining  the  numeri- 
cal values  of  given  angles. 

Ex.  8.  Any  pair  of  adjacent  angles  of  a parallelogram  is  supple- 
mentary. 

Ex.  9.  If  one  angle  of  a parallelogram  is  a right  angle  the  figure  is 
a rectangle. 

Ex.  10.  The  bisectors  of  two  supplementary 
adjacent  angles  form  a right  angle  (are  perpen- 
dicular). 

Ex.  11.  The  bisectors  of  two  interior  angles  on  the  same  side  of  a 
- transversal  to  two  parallel  lines 

form  a right  angle.  2) 

Ex.  12.  If  one  of  the  legs  / j 

(AB)  of  an  isosceles  triangle  / j 

be  produced  its  own  length  ( BD ) and  its  ex- 
tremity ( D ) be  joined  to  the  other  end  of  the  base 
(C),  the  line  last  drawn  {DC)  is  perpendicular  to 
Ine  base.  A, 

193.  Algebraic  method  of  proving  theorems.  The  proof 
cf  certain  properties  of  a geometric  figure  is  often  facilitated 
by  the  use  of  an  algebraic  symbol  for  an  unknown  angle  or 
<m  unknown  line  of  the  figure,  and  the  use  of  an  equation  or 
other  algebraic  method  of  solution , 


92 


BOOK  I.  PLANE  GEOMETRY 


EXERCISES.  CROUP  9 

ALGEBRAIC  METHOD 

Ex.  1.  Find  the  number  of  degrees  in  an  angle  which,  equals  twice 
its  complement? 

[Sug.  Let  a:=the  complement,  etc.] 

Ex.  2.  Find  the  number  of  degrees  in  an  angle  which  equals  its ' 
supplement?  in  one  which  equals  one-third  its  supplement? 

Ex.  3.  The  angular  space  about  a point  is  divided  into  four  angles 
which  are  in  the  ratio  1,  2,  3,  4.  Find  the  number  of  degrees  in  each 
angle. 

[Sug.  x -(-  2 a:  -f-  3 x + 4 #=360°,  etc.]. 

Ex.  4.  The  angles  of  a triangle  are  in  the  ratio  1,  2,  3 ; find  the 
angles. 

Ex.  5.  Two  angles  are  supplementary  and  the  greater  exceeds  the 
less  by  30°;  find  the  angles. 

Ex.  6.  Find  all  the  angles  of  a parallelogram  if  one  of  them  is 
double  another  angle. 

Ex.  7.  One  of  the  base  angles  of  a triangle  is  double  the  other,  and 
the  exterior  angle  at  tbe  vertex  is  105°.  Find  the  angles  of  the  triangle. 

Ex.  8.  How  many  sides  has  a polygon  the  sum  of  whose  angles  is 
fourteen  right  angles? 

[Sug.  2 ( n — 2 ) = 14 ; find  «.] 

Ex.  9.  How  many  sides  has  a polygon  the  sum  of  whose  angles  is 
ten  right  angles?  twenty  right  angles?  720°? 

Ex.  10.  How  many  sides  has  an  equiangular  polygon  one  of  whose 
angles  is  seven -fourths  of  a right  angle  ? 

Ex.  11.  How  many  sides  has  a polygon  the  sum  of  whose  interior!- 
angles  equals  the  sum  of  the  exterior  angles  ? 

Ex.  12.  How  many  sides  has  a polygon  the  sum  of  whose  interior 
angles  equals  three  times  the  sum  of  the  exterior  angles? 

Ex.  13.  If  the  base  of  any  triangle  be  produced  in  both  directions, 
the  sum  of  the  exterior  angles  thus  formed,  diminished 
by  the  vertex  angle,  is  equal  to  two  right  angles. 

[Sue.  180°—  a + 180°-~fc~  (18Cc— a—  &)  = - etc.] 


EXERCISES.  AUXILIARY  LINES 


93 


Ex.  14.  Tlie  bisectors  of  tbo  base  angles  of  an 
isosceles  triangle  include  an  angle  which  is  equal  to 
the  exterior  angle  at  the  base. 

[Sug.  To  prove  a=b,  denote  one  of  the  base  A by 
2 x , etc.] 


Ex.  15.  In  an  isosceles  triangle  the  altitude  upon 
one  of  the  legs  makes  an  angle  with  the  base  which 
equals  one-half  the  vertex  angle. 

[Sug.  To  prove  a=i  b,  show  that  a = 90° — x,  Z>  = 1S0° 
— 2x,  etc.] 


Ex.  16.  If  the  opposite  angles  of  a quadri- 
lateral are  equal,  the  figure  is  a parallelogram. 
[Sug.  2x  + 2i/  = 360°,  etc.] 


194.  Use  of  auxiliary  lines.  Inequalities.  Tlie  demon- 
stration of  a property  of  a geometrical  figure  is  frequently 
facilitated  by  drawing  one  or  more  auxiliary  lines  on  the 
figure.  For  examples  of  the  use  of  such  lines,  see  Props. 
Ill,  V,  IX,  etc.,  of  Book  I. 

Some  of  the  principal  auxiliary  lines  used  on  rectilinear 
figures  are: 

a line  connecting  two  given  points; 
a line  through  a given  point  parallel  to  a given  line; 
a line  through  a given  point  perpendicular  to  a given  line ; 
a line  making  a given  angle  with  a given  line; 

« line  produced  its  own  length,  etc. 


EXERCISES.  CROUP  IO 


AUXILIARY  LINES 

Ex.  1.  In  the  quadrilateral  ABCD  given 
AB=AD,  and  BC=CD-,  prove  AB  = AD. 
[Sug.  Draw  AC,  etc.] 

Ex.  2.  Prove  that  the  angles  at  the  base 
of  an  isosceles  trapezoid  are  equal. 


94 


BOOK  I.  PLANE  GEOMETBY 


Ex.  3.  State  and  prove  the  converse  of  Ex.  2. 


Ex.  4.  Given  AB  ||  CD ; prove  Z & = Z a + Z c 

Ex.  5.  Conversely,  given  Z b , 

= Za-fZc;  prove  AB  ||  CD.  / 


Ex.  6.  The  median  to  the  hypotenuse  of  a right 
triangle  is  one-half  the  hypotenuse. 


Ex.  7.  If  one  acute  angle  of  a right  triangle  is  double 
the  other,  the  hypotenuse  is  double  the  shorter  leg. 
[Sug.  Draw  the  median  to  the  hypotenuse,  etc.] 


Ex.  8.  In  an  isosceles  triangle, 
the  sum  of  the  perpendiculars 
drawn  from  any  point  in  the  base 
to  the  legs  is  equal  to  the  altitude  upon  one  of 
the  legs. 


In  some  cases  it  is  useful  to  draw  two  or  more  auxiliary 
lines. 


Ex.  9.  Show  that  the  median  of  a trapezoid  equals  one-half  the 
sum  of  the  two  bases  by  drawing  a line  through  the  midpoint  of  one 
leg  of  the  trapezoid,  parallel  to  the  other  leg  and  meeting  one  base 
and  the  other  base  produced. 


Ex.  10.  Lines  joining  the  midpoints  of  the  sides  of  a quadrilateral 
taken  in  order  form  a parallelogram. 


[Sug.  Draw  the  diagonals 
eral  and  use  Art.  177.] 

Ex.  11.  If  the  opposite 
sides  of  a hexagon  are  equal 
and  one  pair  of  sides  [AB  and 
CD)  are  parallel,  the  opposite  angles  of  the  hexagon 
are  equal. 

Ex.  12.  Given  AB^BC,  and  AD—C£',  prove 
PF=F£, 


of  the  quadrilat- 
B 


£ 


EXERCISES.  INEQUALITIES 


95 


Let  the  student  form  a list  of  the  'principles  proved  in 
Book  I concerning  unequal  lines  and  unequal  angles. 

(See  Arts.  92,  93,  95,  etc.). 

EXERCISES.  CROUP  II 

INEQUALITIES 

Ex.  1.  in  Uie  triangle  ABC  let  P be  any  point  in  the  side  BC; 
prove  AB  -\-  BO  > AP  -f-  PC. 

Ex.  2.  In  the  quadrilateral  ABCD  let  F be  any  point  in  the  side 
BC;  prove  perimeter  of  ABCD  > perimeter  of  AFD. 

Ex  3.  In  the  triangle  ABC  let  D be  any  point  in  AB  and  F any 
point  in  BC.  Prove  AB  + BC  > AD  -f-  DF  + FC. 

Ex.  4.  The  sum  of  the  four  sides  of  a quadrilateral  is  greater  than 
the  sum  of  the  diagonals. 

Ex.  5.  If,  from  any  point  within  a triangle,  lines  be  drawn  to  the 
vertices,  the  sum  of  the  lines  drawn  is  greater  than  one -half  the  sum 
of  the  sides  of  the  triangle. 

[Sue.  Use  Art.  92  three  times,  etc.] 

Ex.  6.  If,  from  any  point,  within  a triangle,  lines  be  drawn  to  the 
vertices,  the  sum  of  the  lines  drawn  is  less  than  the  sum  of  the  sides 
of  the  triangle. 

[Sug.  Use  Art.  95.] 

Ex.  7.  The  median  to  any  side  of  a triangle  is 
less  than  half  the  sum  of  the  other  two  sides. 

[Sug.  Produce  the  median  its  own  length.] 

Ex.  8.  If  B is  the  vertex  of  an  isosceles  triangle  ABC,  and  BC  be 
produced  to  the  point  D,  /.BAD  is  greater  than  /BDA. 

Ex  9 In  the  figure  p.  71,  BC  > AB;  prove  /BFC  greater  than 
/BFA.  [Sug.  Use  Art.  108.] 

Ex.  10  In  the  same  figure,  show  that  FC  < BC. 

Ex.  11.  In  the  quadrilateral  ABCD,  AD  is  the  longest  side  and 
BC  the  shortest;  prove  / ABC  greater  than  / ADC,  and  / BCD  greater 
than  /BAD. 

Ex.  12  Lines  are  drawn  from  A,  B,  C,  D,  four  points  in  a straight 
line,  to  the  point  F outside  of  the  line  Which  angles  on  the  figure 
are  less  than  angle  ACF%  Which  angles  are  greater  ? 


96 


BOOK  I.  PLANE  GEOMETRY 


195.  Indirect  demonstrations.  Loci.  In  Book  I three 
methods  of  indirect  proof  have  been  used. 

1.  The  reduction  to  an  absurdity  (reductio  ad  absur- 
dum) , that  is,  the  proof  that  the  negative  of  a given  theorem 
leads  to  an  absurdity  (see  Prop.  X). 

2.  The  method  of  exclusion,  that  is,  showing  that  any 
other  statement  than  the  given  theorem  cannot  be  true  (see 
Props.  XV,  XVII).  This  method  is  a special  case  of  the 
preceding,  the  negative  of  a given  theorem  being  divided 
in  it  into  two  parts  which  are  separately  shown  to  be 
impossible. 

3.  The  method  of  coincidence,  that  is,  proof  that  a given 
line  coincides  with  another  line,  which  fulfils  certain  re- 
quired conditions  (see  Props.  XVII,  XXIII,  XXV,  etc.). 

EXERCISES.  CROUP  12 

INDIRECT,  OR  NEGATIVE  DEMONSTRATIONS 

Prove  the  following  by  an  indirect  method: 

Ex.  1.  Every  noint  within  an  angle  and  not  in  the  bisector  of  the 
angle  is  unequally  distant  from  the  sides  of  the  angle. 

[Sug.  In  the  given  angle  take  P any  point  not  in  the  bisector  of 
the  angle.  Then,  if  P is  not  unequally  distant  from  AO  and  OB,  it 
must  be  equally  distant  from  them,  etc.] 

Ex.  2.  If  two  straight  lines  are  cut  by  a transversal,  making  the 
alternate  interior  angles  unequal,  the  lines  are  not  parallel. 

Ex.  3.  The  line  joining  the  midpoints  of  two  sides  of  a triangle 
is  parallel  to  the  third  side. 

[Sug.  Through  one  of  the  midpoints  draw  a line  |]  to  the  third 
side,  show  that  it  bisects  the  second  side  and  that  the  line  joining  the 
. midpoints  coincides  with  it.) 


EXERCISES.  LOCI 


97 


Ex.  4.  If,  from  a point  P in  a line  AB,  lines  PC  and  PD  be  drawn 
on  opposite  sides  of  AB  making  the  angle  APC  equal  to  the  angle  BPD, 
PC  and  PD  are  in  the  same  straight  line. 

[Sug.  From  P draw  PQ  in  the  same  straight  line  with  PC  and 
show  that  PD  coincides  with  it.] 

Ex.  5.  The  bisectors  of  two  vertical  angles  are  in  the  same 
straight  line. 

Ex.  6.  In  the  triangle  ABC,  D is  any  point  in  the  side  AB,  and  E is 
any  point  in  the  side  AC.  Prove  that  BE  and  DC  cannot  bisect  each 

other. 

EXERCISES.  CROUP  13 

LOCI 

Ex.  1.  What  is  the  locus  of  all  points  at  a given  distance,  a,  from 
a given  line  T Prove  this. 

Ex.  2.  What  is  the  locus  of  all  points  equidistant  from  two  given 
parallel  lines  ? Prove  this. 

By  use  of  known  loci  (see  Arts.  112-118),  prove  the  following: 

Ex.  3.  The  diagonals  of  a rhombus  are  perpendicular  to  each 
other.  [Sug.  See  Art.  113.] 

Ex.  4.  The  median  of  an  isosceles  triangle  is  perpendicular  to  the 
base. 

Ex.  5.  The  line  that  joins  the  vertices  of  two  isosceles  triangles  on 
the  same  base  is  perpendicular  to  the  base. 

196.  General  method  of  obtaining  a demonstration  of  a 
theorem.  Analysis. 

A clue  to  the  solution  of  some  of  the  more  difficult 
theorems  is  often  obtained  by  proceeding  thus: 

Assume  the  proposed  theorem  as  true;  observe  ivliat  other 
relation  among  the  parts  of  the  figure  must  then  be  true; 
proceed  backward  thus,  step  by  step,  till  the  required,  theorem 
is  found  to  depend  on  some  known  truth;  then,  starting  with 
this  knoivn  truth,  reverse  the  steps  taken,  and  tints  build  up 
a direct  proof  of  the  required  theorem. 

This  method  is  called  solution  by  analysis. 


i 


G 


98 


BOOK  I.  PLANE  GEOMETRY 


The  following  is  a simple  example  of  the  use  of  this 
method: 

Ex.  Given  AB  and  AC  the  legs  of  an  isosceles 
triangle  and  D any  point  on  AB  • prove  DC  greater 
«Uan  DB. 

Analysis.  If  DC  > DB, 

A B is  greater  than  Z DCB . Art.  104. 

Hence  substituting  for  AB  its  equal,  AACB,  we 
have  Z A CB  is  greater  than  ADCB.  Ax.  8. 

But  we  know  that  AACB  > A DCB. 

Hence,  Direct  Proof  (or  Synthesis) 

AACB  is  greater  than  A DCB, 

Z B is  greater  than  Z DCB. 

/.  DC  > DB. 

The  first  part  (analysis)  of  the  above  process  is  to  be 
purely  mental  work  on  the  part  of  the  pupil,  in  investiga- 
ting a given  theorem;  the  second  part  (the  direct  proof,  or 
synthesis)  is  to  be  written  out  as  the  required  solution. 

In  working  the  following  exercises,  this  method  will  be 
found  to  be  necessary  in  the  solution  of  only  a few  of  the 
more  difficult  theorems. 

EXERCISES.  CROUP  14 

THEOREMS  PROVED  BY  VARIOUS  METHODS 

Ex.  1.  If  two  opposite  angles  of  a quadrilateral  are  bisected  by  the 
diagonal  connecting  their  vertices,  the  quadrilateral  is  bisected  by 
this  diagonal. 

Ex.  2 Perpendiculars  drawn  from  the  extremities  of  the  base  of  a 
triangle  to  the  median  to  the  base,  are  equal. 

Ex.  3.  If  the  perpendiculars  from  the  extremities  of  the  base  of  a 
triangle  to  the  other  two  sides  are  equal,  (1)  these  perpendiculars  make 
equal  angles  with  the  base,  (2)  the  triangle  is  isosceles. 


A 


Ax.  7. 

Ax.  7. 
Ax.  8. 

Art.  106. 

Q.  E.  D. 


MISCELLANEOUS  EXERCISES 


99 


Ex.  4.  If  the  lines  AB  and  CD  intersect,  then  AB-\-CD  > AC-\-DB. 

Ex.  5.  The  vertex  angle  of  an  isosceles  triangle  is  44°,  and  one  of 
the  base  angles  is  bisected  by  a line  produced  to  meet  the  opposite 
side.  Find  all  the  angles  of  the  figure. 

Ex.  6 In  the  figure  of  Prop.  V prove  that  AAPC  is  greater  than 
A ABC  Also  prove  the  same  in  another  way  by  means  of  an  auxiliary 
line  drawn  through  B and  P. 

Ex.  7.  Perpendiculars  drawn  from  the  midpoint  of  the  base  of  aru 
isosceles  triangle  to  the  legs  are  equal. 

Ex.  8 State  and  prove  the  converse  of  Ex.  7. 

Ex.  9.  In  an  isosceles  triangle  an  exterior  angle  at  the  base  equals 
a right  angle  increased  by  one-half  the  vertical  angle. 

Ex  10  In  a re  entrant  quadrilateral  the  exterior 
angle  at  the  re  entrant  vertex  equals  the  sum  of  the 
three  opposite  interior  angles  ( Ad  — A a + Ab-\-  Ac) . 

[Sue  Draw  an  auxiliary  line  ] 

Ex  11  In  the  equilateral  triangle  ARC,  BC  is  pro- 
duced to  D , prove  BD  > AD  > AB. 

Ex.  12.  If  from  a point  in  the  bisector  of  an  oblique  angle 
lines  are  drawn  parallel  to  the  sides  of  the  angle,  the  quadrilateral 
formed  is  a rhombus 

Ex.  13.  If  two  angles  of  a quadrilateral  are  supplementary,  the 
other  two  angles  are  supplementary. 

Ex.  14.  If  the  median  of  a triangle  is  perpendicular  to  the  base, 
the  triangle  is  isosceles. 

Ex.  15.  Given  AB  — AC. 

ABAC  z=4AB, 
and  DF  JL  BC, 
prove  A EFA  equilateral. 

[Analysis.  If  A FAE  is  equilateral,  AFEA  = 60°:  A DEB  = 60° ; 

S = i 0°.  Hence,  to  get  a direct  proof,  show  that  AC—  30°  by 
using  A bAC  = 4 A C.] 


100 


BOOK  I.  PLANE  GEOMETRY 


Ex.  16.  If  the  diagonals  of  a quadrilateral  bisect  each  other  at 
right  angles,  what  kind  of  a figure  is  the  quadrilateral?  Prove  this. 

Ex.  17.  From  the  point  in  which  the  altitudes  drawn  to  the  legs  of 
an  isosceles  triangle  intersect,  a line  is  drawn  to  the  vertex.  Prove 
that  this  line  bisects  the  angle  at  the  vertex. 

Ex.  18.  If  from  a point  within  an  acute  angle  perpendiculars  are 
drawn  to  the  sides  of  the  angle,  the  angle  formed  by  these  perpendicu- 
lars is  the  supplement  of  the  given  angle. 

Ex.  19.  Lines  joining  the  midpoints  o 
sides  of  a triangle  divide  the  triangle  into 
equal  triangles. 

Ex.  20.  If,  in  the  parallelogram  ABCD, 

BP  = DQ,  then  AQCP  is  a parallelogram. 

[Analysis.  If  AQCP  is  a CU  , MP=and 
||  QC.  .'.  begin  the  direct  proof  by  show- 
ing that  AP  = and  is  )|  QC.~\ 

Ex.  21.  If  the  diagonals  of  a parallelogram  are  equal,  what  kind 
of  a figure  is  the  parallelogram?  Prove  this. 

Ex.  22.  If  the  angle  A of  the  triangle  ABC  is  50°  and  the  exterior 
angle  BCD  is  120°,  which  is  the  largest  side  in  the  triangle  ? 

Ex.  23.  Two  triangles  ai  a equal  if  two  sides  and  the  median  to 
one  of  these  sides  in  one  triangle  are  equal,  respectively,  to  two 
homologous  sides  and  a median  in  the  other. 

Ex.  24.  Two  isosceles  triangles  are  equal  if  the  base  and  an  angle 
cf  one  are  equal  to  the  base  and  the  homologous  angle  of  the  other. 

Ex.  25.  Two  equilateral  triangles  are  equal  if  an  altitude  of  one 
is  equal  to  an  altitude  of  the  other. 

Ex.  26.  If  two  medians  of  a triangle  are  equal,  the  triangle  is 
isosceles. 

[Sug.  On  Fig.  to  Prop.  XL'S7!,  taking  AD=EC,  prove  A AOC  isos- 
celes, AAEC=AADC,  etc.  How  could  this  theorem  be  investigated 
by  analysis  ? ] 


MISCELLANEOUS  EXERCISES 


101 


Ex.  27.  Prove  the  sum  of  the  angles  of  a triangle  equal  to  two 
right  angles  by  drawing  a line  through  the  vertex  of  the  triangle 
parallel  to  the  base. 

Ex.  28.  The  homologous  medians  of  two  equal  triangles  ar6 
equal. 

Ex.  29.  The  bisectors  of  an.  angle  of  a tri- 
angle and  of  the  two  exterior  angles  at  the  other 
vertices  are  concurrent. 

Ex.  30.  If  ARC  is  an  equilateral  triangle  and 
AP  = BQ=  CB,  then  PQB  is  an  equilateral 
triangle. 

Ex.  31.  If  the  two  base  angles  of  a triangle 
be  bisected,  and  through  the  point  of  intersec- 
tion of  the  two  bisectors  a line  be  drawn  parallel 
to  the  base,  the  part  of  this  line  intercepted 
between  the  two  sides  equals  the  sum  of  the 
segments  of  the  sides  included  between  the  par- 
allel and  the  base  (i.  e.,  prove  PQ=AP- (-  QC). 

Ex.  32.  Two  quadrilaterals  are  equal,  if  three  sides  and  the  two 
included  angles  of  one  are  equal  to  three  sides  and  the  two  included 
angles  of  the  other,  respectively. 

Ex.  33.  If  the  diagonals  of  a quadrilateral  bisect  each  other,  the 
figure  is  a parallelogram. 

Ex.  34.  Lines  joining  the  midpoints  of  the  sides  of  a rectangle 
in  order  form  a rhombus. 

Ex.  35.  Lines  joining  the  midpoints  of  the  sides  of  a rhombus 
form  a rectangle. 

Ex.  36.  The  bisectors  of  the  angles  of  a parallelogram  form  a 
rectangle. 

Ex.  37.  The  bisectors  of  the  angles  of  a rectangle  form  a square. 

Ex.  38.  If  lines  be  drawn  through  the  vertices  of  a quadrilateral 
parallel  to  the  diagonals,  a parallelogram  is  formed  which  is  twice 
as  large  as  the  original  quadrilateral. 


102 


BOOK  I.  PLANE  GEOMETET 


Ex.  39.  Lines  drawn  from  two  opposite  vertices  of  a parallelo- 
gram to  the  midpoints  of  a pair  of  opposite  sides  trisect  a diagonal 
of  the  parallelogram. 

Ex.  40.  On  the  diagonal  AC  of  a parallelogram  ABCD  equal  parts, 
AP  and  CQ,  are  marked  off.  Prove  BPDQ  a parallelogram.  How 
many  pairs  of  equal  triangles  does  the  figure  contain  f 

Ex.  41.  The  opposite  angles  of  an  isosceles  trapezoid  are  supple 
mentary. 

Ex.  42.  In  an  isosceles  trapezoid,  the  diagonals  are  equal. 

Ex.  43.  If  the  upper  base  of  an  isosceles  trapezoid  equals  the  sum 
of  the  legs,  and  lines  be  drawn  from  the  midpoint  of  the  upper  base  to 
the  extremities  of  the  lower  base,  how  many  isosceles  triangles  are 
formed  ? Prove  this. 

Ex.  44.  The  lines  joining  the  midpoints  of  the  sides  of  an  isos- 
celes trapezoid,  taken  in  order,  form  a rhombus  or  a square. 

Ex.  45.  The  bisectors  of  the  angles  of  a trapezoid  form  a quadri- 
lateral whose  opposite  angles  are  supplementary. 


Book  II 

THE  CIRCLE 


197.  A circle  is  a portion  of  a plane  bounded  by  a 
curved  line,  all  points  of  which  are  equally  distant  from  a 
point  within  called  the  center. 

The  circumference  of  a circle  is  the  curved  line  bounding 
the  circle.  The  term  circle  may  also  be  used  for  the 
bounding  line,  if  no  ambiguity  results. 

A circle  is  named  by  naming  its  center,  as  the  circle  0 ; or  by 
naming  two  or  more  points  on  its  circumference,  as  the  circle  ACD. 


198.  A radius  of  a circle  is  a 
straight  line  drawn  from  the  center 
to  any  point  on  the  circumference, 
as  AO.  A diameter  of  a circle  is  a 
straight  line  drawn  through  the  cen- 
ter and  terminated  by  the  circumfe- 
rence, as  BC. 


Fig.  i 


199.  An  arc  is  any  portion  of 
a circumference,  as  AO.  A semi- 
circumference is  an  arc  equal  to 
one-half  the  circumference,  as 
BAG.  A quadrant  is  an  arc  equal 
to  one -fourth  of  a circumference, 
as  BD. 

200.  A chord  is  a straight  line 
joining  the  extremities  of  an  arc, 
as  EF . 


(103) 


104 


BOOK  II.  PLANE  GEOMETEY 


Every  chord  subtends  two  arcs.  A minor  arc  is  the 
smaller  of  two  arcs  subtended  by  a chord.  A major  arc  is 
the  larger  of  two  arcs  subtended  by  a chord.  Thus,  for  the 
chord  EE  the  minor  arc  is  EPF , and  the  major  arc  is  ETF. 

Conjugate  arcs  is  a general  term  for  a pair  of  minor  and 
major  arcs. 

If  the  arc  subtended  by  a given  chord  is  mentioned, 
unless  it  is  otherwise  specified,  the  minor  arc  is  meant. 

20  i.  A tangent  to  a circle  is  a straight  line  which,  if 
produced,  has  but  one  point  in  common  with  the  circle,  as 
MN.  Hence,  a tangent  touches  the  circumference  in  one 
point  only. 

A secant  is  a straight  line  which,  if  produced,  intersects 
the  cii*cumference  in  two  points,  as  OH. 


202.  A segment  of  a circle  is  a portion  of  the  circle 
bounded  by  an  arc  and  its  chord,  as  ABC  (Fig.  3). 

Into  how  many  segments  does  each  chord  divide  a circle? 

A semicircle  is  a segment  bounded  by  a semicircumfer- 
ence and  its  diameter. 

203.  A sector  of  a circle  is  a portion  of  a circle  bounded 
by  two  radii  and  the  arc  included  by  them,  as  POQ 
(Fig.  3), 


THE  CIRCLE 


105 


204.  A central  angle  is  an  angle  whose  vertex  is  at  the 
center  and  whose  sides  are  radii,  as  the  angle  POQ 
(Fig.  3). 

An  inscribed  angle  is  an  angle  whose  vertex  is  in  the 
circumference  and  whose  sides  are  chords,  as  the  angle  ABG 
(Fig.  4). 

An  angle  inscribed  in  a segment  is  an  angle  whose  ver- 
tex is  in  the  arc  of  the  segment  and  whose  sides  are  chords 
drawn  from  the  vertex  to  the  extremities  of  the  arc.  Let 
the  pupil  draw  a circle,  a segment  in  it,  and  an  angle  in- 
scribed in  the  segment. 

205.  Two  circles  tangent  to  each  other  are  circles  which 
are  tangent  to  the  same  straight  line  at  the  same  point. 
They  are  tangent  internally  or  externally  according  as  one 
circle  lies  entirely  within  or  eutirely  without  the  other. 
See  the  figures,  page  122. 

Concentric  circles  are  circles  which  have  the  same  center. 
Let  the  pupil  draw  a pair  of  concentric  circles. 

206.  A polygon  inscribed  in  a circle  is  a polygon  all  of 
whose  vertices  lie  in  the  circumference  of  the  circle,  as 
ABCDE  (Fig.  4). 

A circle  circumscribed  about  a polygon  is  a circle  whose 
circumference  passes  through  every  vertex  of  the  polygon. 

207.  A polygon  circumscribed  about  a circle  is  a polygon 
all  of  whose  sides  are  tangent  to  the  circle,  as  PQBST 
(Fig.  5). 

A circle  inscribed  in  a polygon  is  a circle  to  which  all 
the  sides  of  the. polygon  are  tangent. 

Concyclic  points  are  points  lying  on  the  same  circum- 
ference. 


106- 


book  II.  PLANE  GEOMETKY 


PROPERTIES  OF  THE  CIRCLE  INFERRED  IMMEDIATELY 

208.  Radii  of  the  same  circle,  or  of  equal  circles,  are 
equal. 

209.  The  diameter  of  a circle  equals  twice  its  radius. 

210.  Diameters  of  the  same  circle,  or  of  equal  circles, 
are  equal. 

211.  If  two  circles  are  equal  (i.  e.,  may  he  made  to 
coincide,  Art.  13),  their  radii  are  equal,  and  conversely . 

212.  A diameter  of  a circle  bisects  the  circle.  For,  by 
placing  the  two  parts  of  the  circle  so  that  the  diameters 
coincide  and  their  arcs  fall  on  the  same  side  of  the  diame- 
ter, these  arcs  will  coincide  (Art.  197). 

213.  A straight  line  cannot  intersect  a circle  in  more 
than  two  points.  For,  if  a straight  line  can  intersect  a 
circle  in  three  (or  more)  points,  three  or  more  equal  lines 
(radii)  can  be  drawn  from  the  same  point  (the  center)  to 
the  straight  line.  But  this  is  impossible  (Art.  111). 


Proposition  I.  Theorem 


214. 

chord. 


A diameter  of  a circle  is  longer  than  any  other 


Given  AB  a diameter,  and  CD  any  other  chord  m the 
circle  0 
To  prove 


AB  > CD. 


THE  CIECLE 


107 


Proof.  Draw  the  radii  OC  and  OD. 

Then,  in  the  A OCD,  0(7+  OD>  CD.  (Why?) 

Substituting  for  OC  its  equal  OA,  and  for  OD  its  equal 
OD,  Ax.  8. 

OA  + OB,  or  AB  > CD. 

Q.  E.  J5. 

Proposition  II.  Theorem 

215.  In  the  same  circle,  or  in  equal  circles,  equal  cen- 
tral angles  intercept  equal  arcs  on  the  circumference. 


i Given  the  equal  circles  0 and  O',  and  the  equal  central 
A AOB  and  A'O'B'. 

To  prove  the  are  AB  = arc  A'B'. 

Proof.  Apply  the  circle  O'  to  the  circle  0 so  that  the 
center  O'  coincides  with  the  center  0,  and  the  radius  O' A' 
with  the  radius  OA. 

Then  the  radius  O'B'  will  fall  on  the  radius  OB, 
{for  by  hyp.  it  AOB  = Z A'O’B’). 

And  B'  will  fall  on  B,  Art.  208. 

( for  OB  = O'B',  being  radii  of  equal  ©). 

Hence  arc  A'B'  will  coincide  with  AB,  Art.  197. 
( for  all  points  of  each  arc  are  equidistant  from  the  center). 

: arc  A'B'  — arc  AB.  Art.  47. 

Q.  E.  D. 


108 


BOOK  II.  PLANE  GEOMETRY 


Proposition  III.  Theorem  (Converse  of  Prop  II) 

216.  In  the  same  circle , or  in  equal  circles,  equal  arcs 
subtend  equal  angles  at  the  center. 


Given  the  equal  circles  0 and  O',  and  the  equal  arcs 
AB  and  A'B'  subtending  the  central  A 0 and  O'. 

To  prove  AO  = A O' . 

Proof.  Apply  the  circle.  O'  to  the  equal  circle  0 so  that 
the  center  O'  coincides  with  the  center  0 and  the  point  A' 
with  the  point  A. 

Then  the  point  B'  will  fall  on  B, 

( for  arc  A'B' = arc  AB  by  hyp.). 

Hence  the  radius  O' A'  will  coincide  with  OA,  and  radius 
O'B'  with  OB,  Art.  66. 

(between  two  points  only  one  straight  line  can  be  drawn). 

:.  A 0 and  A O'  coincide. 

.-.  AO  = A O'.  Art.  4T. 

0.  E.  D. 

? 217.  Cor.  In  the  same  circle,  or  in  equal  circles,  of 

hvo  unequal  central  angles  the  greater  angle  intercepts  the 
greater  arc,  and,  conversely,  of  two  unequal  arcs,  the  greater- 
arc  subtends  the  greater  angle  at  the  center. 


Ex.  Draw  a circle  and  in  it  a segment  which  is  less  than  the 
sector  having  the  same  arc.  Also  one  that  is  greater.  Also  one  that 
is  equal. 


THE  CIECLE 


109 


Proposition  IV.  Theorem 


218.  In  the  same  circle,  or 'in  equal  circles,  equal  chords 
subtend  equal  arcs. 


Given  the  equal  circles  0 and  O' , and  the  chord  AB  — 
chord  A'B' . 

To  prove  arc  AB  = arc  A'B' . 

Proof.  Draw  the  radii  OA,  OB,  O'A',  O'B' . 

Then,  in  the  A AOB  and  A'O'B' , 


Ex.  1.  In  the  above  figure,  if  chord  AB  = 1 in.,  chord  A'B'=  1 in., 
and  arc  AB=  1}  in.,  find  the  length  of  arc  A'B' 

Ex.  2.  Draw  a circle  and  mark  off  a part  of  it  that  is  both  a seg- 
ment and  a sector. 

Ex.  3.  If  the  distance  from  the  center  of  a circle  to  a line  13 
greater  than  tha  radius,  will  the  line  intersect  the  circumference  ? 


AB  — A'B'. 


(Why?) 
(Why?) 
(Why?) 
(Why 
Art.  215. 


AO=A'  O' , and  B0=B'0'. 
.-.  A AOB=A  A'O'B'. 


.-.  1.0=  Z O'. 
arc  AB  = are  A'B'. 


110 


BOOK  II.  PLANE  GEOMETKY 


Proposition  V.  Theorem 

219.  In  the  same  circle , or  in  equal  circles , equal  arc± 
are  subtended  by  equal  chords. 


Given  the  equal  circles  0 and  O',  and  arc  A.B  = arc  A'B1 . 

To  prove  chord  AB  = chord  A'B' . 

Proof.  Draw  the  radii  AO,  BO,  A'O’ , B'O' . 

Then,  in  the  A AOB  and  A'O'B' , 

/O  = / O',  Art.  216. 

( for  arc  AB — A'B' , and,  in  the  same  O,  or  in  equal  ©,  equal  arcs  sub- 
tend equal  A at  the  center ). 

Also  L OA=  O' A',  and  OB—  O'B'.  (Why?) 

.'.  A AOB=  A A'O'B'.  (Why?) 

.*.  AB  = A'B’.  (Why?) 

Q.  E.  D. 


Ex.  1.  In  the  above  figure  if  are  AB=1$  in.,  arc  A'B'  = li  in.,  and 
chord  AB=  1 in.,  find  chord  A'B'  without  measuring  it. 

Ex.  2.  Draw  two  circles  so  that  the  radius  of  one  is  the  diameter 
of  the  other. 

Ex.  3.  To  which  of  the  classes  of  figures  mentioned  in  Art.  IS 
does  a sector  belong  ? a segment  t 


THE  CIRCLE 


111 


Proposition  VI.  Theorem 

220.  In  the  same  circle , or  in  equal  circles,  the  greater 
of  two  {minor)  arcs  is  subtended  by  the  greater  chord;  and, 
Conversely,  the  greater  of  two  chords  subtends  the  greater 
(minor)  arc. 


Given  the  equal  circles  0 and  O',  and  arc  AB  > arc  DF. 
To  prove  chord  AB  > chord  DF. 

Proof.  Draw  the  radii  OA,  OB,  O'D,  O'F. 

Then,  in  the  A AOB  and  DO'F, 

OA  = OfD,  and  OB  = O'F.  (Why?) 

Z 0 is  greater  than  Z O',  Art.  217. 

( for  arc  AB  > arc  DF,  and,  in  the  same  O,  or  in  equal  ©,  of  two  un- 
equal arcs  the  greater  arc  subtends  the  greater  angle  at  the  center ) . 

.'.  chord  AB  > chord  DF,  Art.  107. 

(if  two  A have  txco  sides  of  one  equal  to  two  sides  of  the  other,  but  the 
included  angle  of  the  first  greater,  etc.). 

Conversely.  Given  the  equal  circles  0 and  O',  and 
chord  AB  > chord  DF. 

To  prove  arc  AB  > arc  DF. 

Proof.  In  the  A AOB  and  DO'F, 

OA  = O'D,  and  OB  = O'F.  (Why  ?) 

AB  > DF.  (Why  ?) 

.*.  Z 0 is  greater  than  Z O'.  (Why  ?) 

.*.  arc  AB  > arc  DF,  Art.  217. 

(in  the  same  G,  or  in  equal  ©,  of  two  unequal  central  A the  greater 
angle  intercepts  the  greater  arc). 

6,  £.  D. 


112 


BOOK  II.  PLANE  GEOMETRY 


Proposition  VII.  Theorem 

221.  A diameter  perpendicular  to  a chord  bisects  the 
chord  and  the  arcs  subtended  by  the  chord. 


Given  the  circle  0,  and  the  diameter  PQ  _L  chord  AB. 
To  prove  that  PQ  bisects  the  chord  AB  and  the  arcs  APB 
and  AQB. 

Proof.  Draw  the  radii  OA  and  OB. 

Then,  in  the  rt.  A OAR  and  OBR, 


OA  = OB. 

(Why?) 

OR  = OR. 

(Why?) 

A OAR  = A OBR. 

(Why?) 

:.  AR  — BR , and  Z AOR  = ZBOR. 

(Why  5) 

arc  A P = arc  BP, 

Art,  215. 

(in  the  same  Q,orin  = ©,  = central  A intercept  = arcs  on  the  circf.) 

ZAOQ  = ZBOQ  (Art.  75).  .-.  arc  AQ  = arc  BQ.  (Why  ?) 

Q.  E.  D. 

222.  Cor.  1.  A diameter  which  bisects  a chord  (shorter 
i loan  a diameter)  is  perpendicular  to  the  chord ; a diameter 

which  bisects  an  arc  is  the  perpendicular  bisector  of  that  arc. 

223.  Cor.  2.  The  perpendicular  bisector  of  a chord 
passes  through  the  center  of  the  circle,  and  bisects  the  arcs 
subtended  by  the  chord. 

224.  Cor.  3.  A line  from  the  center  perpendicular  to 
a chord  bisects  the  chord. 

225.  Cor.  4.  A line  passing  through  the  midpoints  of 
a chord  and  its  arc  passes  through  the  center,  and  is  a diame- 
ter perpendicular  to  the  chord. 


THE  CIRCLE 


j.i3 

Proposition  VIII.  Theorem 

226.  In  the  same,  circle , or  in  equal  circles,  equal  chords 
ere  equidistant  from  the  center ; and,  conversely,  chords 
which  are  equidistant  from  the  center  are  equal. 


Given  the  circle  0 in  which  the  chords  AB  and  CD 
are  equal. 

To  prove  that  AB  and  CD  are  equidistant  from  the 
center. 

Proof.  Let  OB  be  drawn  _L  AB,  and  OF  JL  CD. 

Di’aw  the  radii  OA  and  OC. 

Then  OB  bisects  AB,  and  OF  bisects  CD,  Art.  224. 
[a  line  from  the  center  J_  a chord  bisects  the  chord). 

Hence,  in  the  rt.  A OAE  and  OCF, 


OA  = OC.  (Why?) 

AE  = CF.  Ax.  5. 

.*.  AOAE  = A OCF.  (Why?) 

.-.  OF  = OF.  (Why?) 


Conversely.  Given  circle  0,  and  AB  and  CD  equidis- 
tant from  the  center. 

To  prove  AB—CD. 

Proof.  Let  the  pupil  supply  the  proof. 


114 


BOOK  II.  PLANE  GEOMETRY 


Proposition  IX.  Theorem 

227.  In  the  same  circle,  or  in  equal  circles , if  two  chords 
are  unequal,  they  are  unequally  distant  from  the  center,  and 
the  less  chord  is  at  the  greater  distance  from  the  center. 


Given  in  the  circle  0 the  chord  CD  < chord  AB. 

To  prove  that  chord  CD  is  at  a greater  distance  from  the 
center  than  chord  AB. 

Proof.  Let  OG  be  drawn  _L  CD,  and  OF  _L  AB. 

Then  chord  AB  > chord  CD.  Hyp. 

arc  AB  > arc  CD,  Art.  220. 

(in  the  same  O,  or  in  equal  ©,  the  greater  of  two  minor  arcs  is  subtended 
by  the  greater  chord,  and  conversely) . 

Mark  off  on  the  arc  AB  the  arc  AE— arc  CD,  and  draw 
the  chord  AE. 

Chord  AE  = chord  CD,  Art.  219. 

(in  the  same  O,  or  in  equal  ®,  equal  arcs  are  subtended  by  equal  chords). 

Let  OH  be  drawn  J_  AE,  and  intersecting  AB  at  L. 

:.  OH  = OG,  Art,  226. 

(in  the  same  O,  or  in  equal  ©,  equal  chords  are  equidistant  from  the 

center). 

But  OH  > OL.  As.  7. 

Also  OL  > OF.  (Why?) 

Much  more  then  OH,  or  its  equal  OG  > OF.  As.  12. 

9.  E D. 


THE  CIRCLE 


115 


Proposition  X.  Theorem  (Converse  of  Prop.  IX) 

228.  In  the  same  circle , or  in  equal  circles , if  two  chords 
are  unequally  distant  from  the  center,  the  more  remote  is 
the  less. 


Given  in  the  circle  0 the  chord  CD  farther  from  the 
center  than  the  chord  AB. 

To  prove  chord  CD  < chord  AB. 

Proof.  Let  OH  be  drawn  _L  CD,  and  OG  _L  AB. 

OH  > OG.  (Why?) 

On  OH  mark  off  OL  = OG. 

Through  L let  the  chord  ELF  be  drawn  _L  OH. 

Then  chord  EF  — chord  AB,  Art.  226. 

{in  the  same  O,  or  in  equal  ®,  chords  which  are  equidistant  from  the 
center  are  equal ) . 

But  the  arc  CD  < arc  EF.  Ax.  7. 

chord  CD  < chord  EE,  or  its  equal  AB,  Aft.  220. 

{in  the  same  Q,  or  in  equal  ® , the  greater  of  two  minor  arcs  is  subtended 
by  the  greater  chord,  and  conversely) . 


Q.  E.  P. 


116 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XI.  Theorem 

229.  A straight  line  perpendicular  to  a radius  at  its 
extremity  is  tangent  to  the  circle. 


Given,  the  circle  O,  the  radius  OA , and  the  line  BC  1.  OA 
at  its  extremity  A. 

To  prove  that  BC  is  tangent  to  the  O. 

Proof.  Take  P,  any  point  on  the  line  BC  except  A, 
and  draw  OP. 

Then  OP  > OA . (Why?) 

Hence  the  point  P lies  without  the  circle. 

.'.  every  point  in  the  line  BC,  except  A,  lies  outside 
the  O. 

.-.  BC  is  tangent  to  the  circle,  Art.  201. 

(a  tangent  to  a O is  a straight  line  which,  etc.). 

Q.  E.  D. 

230.  Cor.  1.  The  radius  drawn  to  the  point  of  contact'' 
is  perpendicular  to  a tangent  to  a circle. 

231.  Cor.  2.  A perpendicular  to  a tangent  at  the  point 
of  contact  passes  through  the  center  of  the  circle. 

232.  Cor.  3.  The  perpendicular  drawn  from  the  center 
of  a circle  to  a tangent  passes  through  the  point  of  contact . 


117 


THE  CIRCLE 


Proposition  XII.  Theorem 

233.  Two  parallel  lines  intercept  equal  arcs  on  a cir- 
cumference. 


Fig.  t 


Case  I.  Given  AB  (Fig.  1)  tangent  to  the  O PCD  at 
P,  CD  a secant  ||  AB  and  intersecting  the  circumference  in  C 

and  D. 


To  prove  arc  PC  = arc  PD. 

Proof.  Draw  the  diameter  PQ. 

Then  PQ  JL  AB.  Art.  230. 

PQ  J_  CD.  Art.  123. 

.*.  arc  PC  = arc  PD,  Art.  221. 

(a  diameter  J.  chord  bisects  the  chord  and  the  arcs  subtended  by 

the  chord). 


Case  II.  Given  AB  and  CD  (Fig.  2)  ||  secants  inter- 
secting the  circumference  in  A,  B and  C,  D respectively. 


To  prove  arc  AC  — arc  BD. 

Proof.  Let  a tangent  EF  be  drawn  ||  AB  and  touching  the 
circle  at  P. 

Then  EF  ||  CD.  (Why?) 


arc  AP  — arc  BP. 


Case  I. 


arc  CP  = arc  DP.  (Why?) 

arc  AC  = arc  BD.  Ax.  s. 


Then 

Also 

Hence 


118 


BOOK  II.  PLANE  GEOMETRY 


Case  III.  Given  AB  and  CD  ("Fig.  3)J|  tangents  touch- 
ing the  O at  P and  Q respectively. 

To  prove  arc  PEQ  = arc  PFQ. 

Proof.  Let  the  pupil  supply  the  proof. 

234.  Cor.  The  straight  line  whizh  joins  the  points  op 
contact  of  two  parallel  tangents  is  a diameter. 


Proposition  XIII.  Theorem 

235.  Through  three  points , not  in  the  same  straight  line , 
one  circumference,  and  only  one,  can  he  drawn. 


Given  A,  B and  C any  three  points  not  in  the  same 
straight  line. 

To  prove  that  one  circumference,  and  only  one,  can  be 
drawn  through  A,  B and  C. 

Proof.  Draw  the  straight  lines  AB  and  BC,  and  let  A 
be  erected  at  the  midpoints,  D and  E,  of  AB  and  BC 
respectively. 

These  A will  intersect  at  some  point  0,  Art.  122. 

( lines  J_  non-parallel  lines  are  not  ||  ). 

But  0 is  in  the  A bisector  of  AB. 


Const. 


THE  CIRCLE 


119 


0 is  equidistant  from  the  points  A and  B.  Art.  112. 

In  like  manner,  0 is  in  the  _L  bisector  of  BC,  and  is 
equidistant  from  the  points  B and  C. 

Hence  0 is  equidistant  from  the  three  points  A,  B and  0. 

Ax.  1. 

Hence  if  a circumference  be  described  with  0 as  a cen- 
ter and  OA  as  a radius,  it  will  pass  through  A,  B and  G. 

Also  DO  and  BO  intersect  in  but  one  point.  Art.  64. 

Hence  there  is  but  one  center. 

Again,  0 is  equally  distant  from  the  points  A,  B and  C; 
hence  there  is  but  one  radius. 

With  only  one  center  and  only  one  radius,  but  one  cir- 
cumference can  be  described. 

Hence  one  circumference,  and  only  one,  can  be  drawn 
through  the  points  A,  B and  C. 

Q.  E.  D. 

236.  Note.  The  theorem  of  Art.  235  enables  us  to  shrink  or 
economize  a circle  into  three  points ; or  to  expand  any  three  points 
into  a circle. 


Ex.  1.  How  many  circumferences  can  be  passed  through  four 
given  points  in  a plane,  each  circumference  passing  through  three, 
and  only  three,  of  the  given  points  ? 

Ex.  2.  Draw  two  circles  so  that  they  can  have  a common  chord. 

Ex.  3.  Can  two  circles  which  are  tangent  to  each  other  have  a 
common  chord  f 

Ex.  4.  Can  two  circles  which  are  tangent  to  each  other  have  a 
common  secant  ? 

Ex.  5.  Draw  two  circles  which  can  have  neither  a common  chord 
nor  a common  tangent. 

Ex  6.  Is  it  possible  to  draw  two  circles  which  cannot  have  a 
conjmon  secant  t 


120 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XIV.  Theorem 

237.  The  two  tangents  drawn  to  a circle  from  a point 
outside  the  circle  are  equal , and  make  equal  angles  with  a 
line  drawn  from  the  point  to  the  center. 


Given  FA  and  PB  two  tangents  drawn  from  the  point 
P to  the  circle  0. 

To  prove  PA  = PB,  and  AAPO=  ABPO. 

Proof.  Let  the  pupil  supply  the  proof. 

238.  Def.  The  line  of  centers  of  two  circles  is  the 
line  joining  their  centers. 

239.  Def.  Two  circles  which  do  not  meet  may  have 
four  common  tangents. 

A common  internal  tangent  of  two  circles  is  a tangent 
which  cuts  their  line  of  centers. 

240.  Def.  A common  external  tangent  of  two  circles  is 

a tangent  which  does  not  cut  their  line  of  centers. 

Ex.  1 . Iii  the  above  figure,  prove  that  the  line  drawn  to  the  center 
from  the  point  in  which  the  two  tangents  meet  makes  equal  angles 
with  the  radii  to  the  points  of  contact. 

Ex.  2.  Draw  a circle  with  a radius  of  3 in.  and  another  with  a radius 
of  2 in.,  with  their  centers  4 in.  apart.  Will  these  circles  intersect  ? 
If  their  centers  were  6 in.  apart,  would  they  intersect  ? 


THE  CIRCLE 


121 


Proposition  XV.  Theorem 

241.  If  two  circles  intersect,  their  line  of  centers  is  per  = 
pendicular  to  tlieir  common  chord  at  its  middle  point. 


Given  the  circles  0 and  O',  intersecting  at  the  points 
A and  B. 

To  prove  00'  J_  AB  at  its  middle  point. 

Proof.  Draw  the  radii  OA,  OB,  O' A,  O'B. 

Then  OA  = OB,  and  0’A  = 0'B.  (Why?) 

Hence  0 and  O'  are  two  points  each  equidistant  from 
A and  B. 

00'  is  the  _L  bisector  of  AB.  Art.  113. 

Q.  E.  D. 

Ex.  1.  Draw  two  intersecting  circles  and  show  that  the  line  of 
centers  is  less  than  the  sum  of  the  radii. 

Draw  two  circles  in  which  the  line  of  centers 
Ex.  2.  Equals  the  sum  of  the  radii. 

Ex.  3.  Is  greater  than  the  sum  of  the  radii. 

Ex.  4.  Is  less  than  the  sum  of  the  radii. 

Ex.  5.  Is  less  than  the  sum,  but  greater  than  the  difference  of  the 
radii. 

Ex.  6.  State  in  general  terms  the  relative  positions  of  the  circles 
in  the  four  preceding  exercises. 


122 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XVI.  Theorem 

242.  If  two  circles  are  tangent  to  each  other , the  line  of 
centers  passes  through  the  point  cf  contact. 


Given  the  circles  0 and  O'  tangent  to  each  other  at  the 
point  R. 

To  prove  that  the  line  00'  passes  through  R. 

Proof.  At  the  point  R let  PQ,  a tangent  to  the  given 
©,  be  drawn. 

Also  let  AB  be  drawn  A PQ  at  R. 

Then  AB  passes  through  0 and  also  through  O',  Art.  231. 
(a  A to  a tangent  at  the  point  of  contact  passes  through  the  center). 

line  AB  coincides  with  the  line  00’.  Art.  64. 

.'.  00'  passes  through  the  point  R , 

( for  it  coincides  with  AB  which  passes  through  It) . 


How  many  common  internal,  and  how  many  common  external 
tangents  have  two  circles 
Ex.  1 . If  they  touch  externally  ? 

Ex.  2.  If  they  touch  internally  ? 

Ex.  3.  If  they  intersect  ? 

Ex.  4.  If  one  circle  lies  wholly  within  the  other  ? 

Ex.  &.  If  one  circle  lies  wholly  without  the  other  ? 


THE  CIRCLE 


123 


EXERCISES.  CROUP  IS 


Ex.  1.  The  line  joining  the  center  of  a circle  to  the  midpoint  of  a 
chord  is  perpendicular  to  the  chord. 

Ex.  2.  A,  B,  C and  D are  four  points  taken  in  succession  on  the 
circumference  of  a circle,  and  arc  AB  — arc  CD.  Prove  that  chord 
AC=chord  BD. 

Ex.  3.  Tangents  drawn  at  the  extremities  of  a diameter  are 
parallel. 

Ex.  4.  PA  and  PB  are  tangents  to  a circle  drawn  from  the  point 
P.  0 is  the  center  of  the  circle.  Prove  that  PO  is  the  perpendicular 
bisector  of  the  chord  AB. 

Ex.  5.  If  the  perpendiculars  from  the  center  upon  two  chords  are 
equal,  the  arcs  subtended  by  these  chords  are  equal. 

Ex.  6.  A,  B , C and  D are  points  taken  in  succession  on  a semi-cir- 
cumference, and  are  AC  is  greater  than  arc  BD]  prove  that  chord 
AB>  chord  CD. 


Ex.  7.  State  the  converse  of  the  preceding 
theorem  and  prove  it. 

Ex.  8.  Given  BA,  BQ  and  QB  tangents  of 
the  circle  O ; prove  BQ=BA  + QB. 

Ex.  9.  If  a quadrilateral  be  circumscribed 
about  a circle,  show  that  the  sum  of  one  pair  of 
opposite  sides  equals  the  sum  of  the  other  pair. 

Ex.  10.  If  a hexagon  be  circumscribed  about  a circle,  show  that  the 
sum  of  three  alternate  sides  equals  the  sum  of  the  other  three  sides 

Ex.  11.  If  a polygon  of  2n  sides  be  circumscribed  about  a circle, 
the  sum  of  n alternate  sides  equals  the  sum  of  the  other  n sides. 


Ex.  12  A circumscribed  parallelogram 
is  equilateral . 

Ex.  13.  If  two  circles  are  tangent 
externally,  the  common  internal  tangent 
bisects  the  common  external  tangent  (i.  e., 
prove  PA  = PB) , 


A 


124 


BOOK  H.  PLANE  GEOMETRY 


Ex.  14.  If  two  circles  are  tangent 
(either  externally  or  internally)  tangents 
drawn  to  them  from  any  point  in  the  com- 
mon tangent  are  equal. 


B 


Ex.  15.  Two  circles 
whose  centers  are  0 and 
O'  are  tangent  internally 
at  P.  The  line  PAB  is 


P' 


drawn  intersecting  the  circumferences  at  A and  B. 
Prove  that  OA  and  O'B  are  parallel. 


MEASUREMENT.  RATIO. 


243.  Measurement.  For  many  purposes,  the  most 
advantageous  way  of  dealing  with  a given  magnitude  is  to 
take  a certain  definite  part  of  the  magnitude  as  a unit, 
and  to  determine  the  number  of  times  this  unit  must  be 
taken  in  order  to  make  up  the  given  magnitude.  Ease  and 
precision  in  dealing  with  magnitudes  are  thus  obtained. 

Geometric  magnitudes  thus  far  have  been  treated  as  wholes,  the 
object  being  simply  to  determine  whether  two  given  magnitudes  are 
equal,  or  unequal,  or  to  determine  some  similar  general  relation. 
Hereafter  geometric  magnitudes  will  frequently  be  treated  as  if  com- 
posed of  units. 

To  measure  a given  magnitude  is  to  find  how  many  times 

O O J 

the  given  magnitude  contains  another  magnitude  of  the  same 
kind  taken  as  a unit. 

244.  The  numerical  measure  of  a magnitude  is  the  num- 
ber which  expresses  how  many  times  the  unit  of  measure  is 
contained  in  the  given  magnitude. 

Thus,  when  a boy  says  that  he  is  five  feet  tall,  he  means  that,  if  a fooJ' 
rule  be  applied  to  his  height,  the  foot  rule  will  be  contained  five  times. 

A quantity  is  often  measured  to  best  advantage  by  measuring  a 
related  but  more  accessible  quantity,  which  has  the  same  numerical  meas- 
ure as  the  original  quantity.  This  process  is  indirect  measurement.  Thus, 
the  temperature  of  the  air  is  measured  indirectly  by  measuring  the 
height  of  a column  of  mercury  in  a thermometer  tube.  So  the  number  of 
times  a unit  of  angle  is  contained  in  a given  angle  is  often  ascertained 


MEASUREMENT 


125 


most  readily  by  determining  tlie  number  of  times  a unit  of  arc  is  con- 
tained in  a given  arc. 

245.  The  ratio  of  two  magnitudes  of  the  same  kind  is 
their  relative  magnitude  as  determined  by  the  number  of 
times  one  is  contained  in  the  other.  Hence,  it  is  the  qua- 
tient,  or  indicated  quotient,  of  the  two  magnitudes. 

, . „ „ „ . . 36  in.  36 

Thus,  the  ratio  of  3 ft.  to  1 ft.  7 in.  is  — — — , or  -. 

19  in.  19 

The  use  of  ratio  is  illustrated  by  the  fact  that  several  indicated  quo. 
tients  when  taken  together  may  be  simplified  by  cancellation  before 
final  determination  of  their  value  is  made. 

Two  magnitudes  of  the  same  kind  have  the  same  ratio  as 
their  numerical  measures. 

246.  Measurement  as  a ratio.  An  important  particular 
instance  of  ratio  is  that  ratio  in  which  one  of  the  two  mag- 
nitudes compared  is  a unit  of  measurement.  Hence,  the 
numerical  measure  of  a magnitude  is  the  ratio  of  the  mag- 
nitude to  the  unit  of  measure. 

Thus,  the  numerical  measure  of  the  height  of  a boy  is  the  ratio  of 
his  height  (5  ft.)  to  the  unit  of  measure  (1  ft.),  or  5. 

247.  Commensurable  magnitudes  are  magnitudes  of  the 
same  kind  which  have  a common  unit  of  measure. 

Thus,  12  ft.  and  25  ft.  have  the  common  unit  of  measure,  1 ft.,  and 
hence  are  commensurable  magnitudes  ; also  13V  bus.  and  7J  bus.  have 
a common  unit,  1 peck,  and  are  commensurable. 

248.  Incommensurable  magnitudes  are  magnitudes  of 
the  same  kind  which  have  no  common  unit  of  measure. 

Ills.  $5  and  $-^3  ; 7 yrs.  and  15  yrs.;  so  the  side  and  the  diagonal 
of  a square  may  be  proved  to  have  no  common  unit  of  measure  ; likewise 
the  diameter  and  the  circumference  of  a circle. 

In  general,  a ratio  which  is  expressed  by  a surd  number, 
as\/2  or  \/3,  is  a ratio  between  incommensurable  magnitudes 
(called  au  incommensurable  ratio). 


126 


BOOK  II.  PLANE  GEOMETRY 


METHOD  OF  LIMITS 

249.  A variable  quantity,  or  a variable,  is  a quantity 
which  may  have  an  indefinite  number  of  different  values 
under  the  conditions  of  a theorem  or  problem. 

Thus,  the  distance  a railroad  train  goes  varies  with  the  number  of 
hours  which  the  train  travels. 

250.  A constant  is  a quantity  which  remains  unchanged 
in  value  in  a given  problem  or  discussion. 

Thus,  if  a polygon  be.  inscribed  in  a circle  and  the  number  of  sides 
of  the  polygon  be  doubled,  quadrupled,  etc.,  the  perimeter  of  the  poly- 
gon will  be  variable,  but  the  circumference  of  the  circle  will  remain 
unchanged,  and  hence  be  a constant. 

25 1 . The  limit  of  a variable  quantity  is  a constant  quantity 
which,  under  the  conditions  of  a given  discussion,  the  given 
variable  may  approach  as  nearly  as  we  please  in  value,  but 
which  the  variable  can  never  equal. 

Thus,  in  the  illustration  of  Art.  250,  the  circumference  of  the  circle 
is  the  limit  of  the  perimeter  of  the  inscribed  polygon  ; also,  the  area  of 
the  circle  is  the  limit  of  the  area  of  the  inscribed  polygon. 

From  the  definition  of  limit,  it  follows  that  the  differ- 
ence between  a variable  and  its  limit  may  be  made  as  small 
as  we  please  but  can  not  become  zero. 

As  another  illustration  of  a variable  and  its  limit,  we  may  take  the 
case  of  a point  P travelling  along  a given  line  AB,  in  such  a way  that 
in  the  first  second  it  passes  over  A Pi,  one-half  the  line  AB]  in  the 
second  second  over  half  the  remaining  part  of  the  line,  and  arrives  at 
P2;  in  the  third  second  over  one- 

half  the  remainder  of  the  line,  and  , , , , , 

arrives  at  P3,  etc.  It  is  evident  .A  ^2  ^3 

that  the  point  P can  never  arrive  at 

B;  for,  in  order  to  do  this,  in  some  one  second  the  point  would  need 
to  pass  over  the  whole  of  the  remaining  distance. 

In  this  illustration,  AP,  the  distance  traveled  by  the  moving  point,  is 
a variable  (depending  on  the  number  of  seconds),  and  AB  is  its  limit. 

If  the  distance  AB  be  denoted  by  2,  the  distance  traveled  will  be 
denoted  by  1+i  + i + i-t-  . . . . and  will  vary  according  to  the 
number  of  terms  of  the  series  taken  > 


METHOD  OE  LIMITS 


127 


252.  Use  of  variables  and  limits.  Many  of  the  proper- 
ties of  limits  are  the  same  as  the  properties  of  the  variables 
approaching  them.  Hence  a demonstration  of  a difficult 
theorem  may  often  be  obtained  by  first  finding  the  properties 
of  relatively  simple  variables  and  then  transferring  these 
properties  to  their  more  complex  limits. 

253.  Properties  of  variables  and  limits. 

1.  The  limit  of  the  sum  of  a number  of  variables  equals 
the  sum  of  the  limits  of  these  variables.  For,  since  the 
difference  between  each  variable  and  its  limit  may  be  made 
as  small  as  we  please,  the  sum  of  all  these  differences  may 
be  made  as  small  as  we  please  (since  it  is  a finite  number 
of  differences,  with  each  difference  approaching  zero). 

2.  The  limit  of  a times  a variable  equals  a times  the 
limit  of  the  variable,  a being  a constant.  For,  if  the  differ- 
ence between  a variable  and  its  limit  may  be  made  as  small 
as  we  please,  a times  this  difference  may  be  made  as  small 
as  we  please. 

3.  The  limit  of  ^th  part  of  a variable  is  jih  part  of  the 

limit  of  the  variable,  a being  a constant.  For,  if  the 
difference  between  a variable  and  its  limit  may  be  made  as 

small  as  we  please,  ifh  part  of  this  difference  may  be  made 
as  small  as  we  please. 

4.  If  a variable  A:  0.  a times  (a  being  finite)  or  ~ part  of 
the  variable ; and  if  a.  diminish  in  the  one,  or  increase  in 
the  other  process,  the  limit  is  still  zero. 

Ex.  1.  Are  2^  gal.  and  3jt  qt.  commensurable?  If  so,  what  is  the 
common  unit  of  measure  ? 

Ex.  2.  If  c denote  a constant  and  v and  v'  variables,  is  the  value  of 

each  of  the  following  a variable  or  a constant : — , — , v X v',  — ? Illus- 

c v v 

trate. 


128  BOOK  II.  PLANE  GEOMETRY 

Proposition  XVII.  Theorem 

254.  If  two  variables  are  always  equal,  and  each  ap- 
proaches a limit,  their  limits  are  equal. 


Given  AB  the  limit  of  the  variable  AP,  AC  the  limit  of 
the  variable  AQ,  and  AP=AQ  always. 

To  prove  AB=AC. 

Proof.  If  the  limit  AB  does  not  equal  the  limit  AC. 
one  of  these  limits,  as  AC,  must  be  larger  than  the  other. 

Then,  on  AC,  take  AD  equal  to  AB. 

But  AQ  may  have  a value  greater  than  AD.  Art.  251. 

Hence  AQ  would  be  greater  than  AB, 

(for  AQ  > AD  which  = AB). 

.\  AQ  > AP,  Ax.  12. 

( for  AB  > AP). 

But  this  is  contrary  to  the  hypothesis  that  AQ  and  AP 
are  always  equal. 


.'.  AC  cannot  be  greater  than  AB. 


In  like  manner  it  may  be  shown  that  AB  is  not  greater 

than  AC. 


:.  AB=AC. 


Q.  E.  D. 


Ex.  What  method  of  proof  is  used  in  Prop.  XVII  ? 


MEASUREMENT  OF  ANGLES 


129 


Proposition  XVIII.  Theorem 


255.  In  the  same  circle , or  in  equal  circles , two  central 
angles  have  the  same  ratio  as  their  intercepted  arcs. 

Case  I When  the  intercepted  arcs  are  commensurable. 


Given  the  equal  © O'  and  0,  with  the  eentral  A A'O’B'  and 
AOB  intercepting  the  commensurable  arcs  A'B'  and  AB. 

Z A'O'B'  A'B' 

Toprove  TImT'IJ- 

Proof.  Let  arc  PQ  (from  a circle  = 0 and  O')  be  a com- 
mon measure  of  the  arcs  A'B'  and  AB. 

PQ  will  be  contained  in  arc  A'B'  an  exact  number  of 
times,  as  7 times,  and  in  AB  an  exact  number  of  times, 
as  5 times. 

arc  A'B'  7 
arc  AB  5 

From  O'  and  0 draw  radii  to  the  several  points  of 
division  of  the  arcs  A'B'  and  AB. 

Then  the  A A'O'B'  will  be  divided  into  7,  and  the5 
A AOB  into  5 small  angles,  all  equal,  Art.  216. 

(in  the  same  o,  or  in  =®  equal  arcs  subtend  equal  A.  at  the  center). 

A A'O'B’  _7 
5 

A'B' 


Then 


Art.  245. 


Hence 


A AOB 

A A'O'B'  

A AOB  AB 


Art.  245. 


Ax.  1. 


I 


130 


BOOK  II.  PLANE  GEOMETRY 


Case  II.  When  the  intercepted  arcs  are  incommen 

surahle. 


Given  the  equal  ® O'  andO,  with  the  central  A A'O'B'  and 
AOB  intercepting  the  incommensurable  arcs  A’B'  and  AB. 

„ A A'O'B'  A'B' 

To  prove  A AOB  AB  ' 


Proof.  Let  the  arc  AB  be  divided  into  any  number  of 
equal  parts,  and  let  one  of  these  parts  be  applied  to  the 
arc  A'B' . It  will  be  contained  in  A'B'  a certain  number 
of  times, with  an  arc  BB'  as  a remainder. 

Hence  the  arcs  A D and  AB  have  a common  unit  of  meas- 
ure. Coustr. 


A A'OD 


AD 
AB  • 


Case  I. 


* ' A AOB 

If  now  we  let  the  unit  of  measure  be  indefinitely  dimin- 
ished, the  arc  DB\  which  is  less  than  the  unit  of  measure.  ?/ill 
be  indefinitely  diminished. 

Hence  arc  AD  Dave  A'B'  as  a limit,  and  A ADD  A 

^ A'O'B'  as  a limit.  Art.  251. 

A ADD  A A'O'B' 

Hence  — , . becomes  a variable  approaching  ■ 


A AOB 
as  its  limit; 


A AOB 

Art.  253,  3. 


AD  A'B ' 

Also  -ttt  becomes  a variable  approaching  as  its 


AB 


limit. 


AB 
Art.  253,  3. 


MEASUREMENT  OF  ANGLES 


131 


But  the  variable  — / - . ^ !;  = the  variable  alwavs. 

£AOB  AB  Case  I. 

„ ,.  ZA'O'B'  „ ,.  A'B' 

the  limit  = the  limit  , ,,  > Art.  254. 


ZAOB 


AB 


(if  two  variables  are  ahvays  equal,  and  each  approaches  a limit,  their  limits 

are  equal). 

Q.  E.  D. 

256.  Def.  A degree  of  arc  is  one  three  hundred  and 
sixtieth  part  of  the  circumference  of  a circle. 


257.  Cor.  The  number  of  degrees  in  a central  angle 
equals  the  number  of  degrees  in  the  intercepted  arc;  that  is, 
a central  angle  is  measured  by  its  intercepted  arc. 


Ex.  1.  What  is  the  ratio  of  a quadrant  to  a semi-circumferenee  ? 

Ex.  2.  What  is  the  ratio  of  an  angle  of  an  equilateral  triangle  to 
one  of  the  acute  angles  of  an  isosceles  right  triangle  ? 

Ex.  3.  Draw  two  circles  so  that  the  center  of  each  circle  is  on  the 
circumference  of  the  other. 

Ex.  4.  Draw  three  circles  so  that  the  center  of  each  is  on  the  cir- 
cumference of  the  other  two. 

[Sug.  First  draw  an  equilateral  triangle.] 

Ex.  5.  Draw  three  circles  each  of  which  shall  be  tangent  to  the 
other  two. 

Ex.  6.  Draw  two  concentric  circles  and  a line  which  is  a tangent 
to  one  of  these  circles  and  a chord  of  the  other. 

Ex.  7.  Draw  two  concentric  circles  and  a line  which  is  a secant  of 
one  and  a chord  of  the  other. 


132 


BOOK  II.  PLANE  GEOMETRY 


Proposition  SIX.  Theorem 

258.  An  inscribed  angle  is  measured  by  one-half  its 
intercepted  arc. 

A A a 


Case  I.  When  the  center  of  the  circle  lies  in  one  side  of 
the  inscribed  angle. 

Given  Z BAG  (Fig.  1 ) inscribed  in  the  circle  0,  and  AB 
passing  through  the  center  0. 

To  prove  that  ABAC  is  measured  by  £ arc  BC. 


Proof.  Draw  the  radius  OC. 

Then,  in  the  A OAC,  OA  — OC.  (Why?) 

IA  = IC.  (Why?) 

But  ABOC  = ZA-h  ZC.  Art,  135. 

IBOC  = 21  A.  Ax.  8. 

But  Z BOG  is  measured  by  arc  BC,  Art.  257. 

( a central  Z is  measured  by  its  intercepted  arc). 

So.  of  angular  degrees  in  Z BOC='iso.  of  arc  degrees  in  BC. 
Hence  Z A is  measured  by  arc  BC.  Ax.  5. 


Case  II.  When  the  center  of  the  circle  lies  within  the 
inscribed  angle. 

Given  the  inscribed  /.BAG  (Fig.  2),  with  the  center 
of  the  circle  0 lying  within  the  angle. 

To  prove  that  ABAC  is  measured  by  £ arc  BC. 

Proof.  Draw  the  diameter  AB. 

Then  Z BAB  is  measured  by  £ arc  BB. 

Let,  the  pupil  complete  the  proof. 


Case  I, 


MEASUREMENT  OF  ANGLES 


133 


Case  III.  When  the  center  of  the  circle  is  without  the 
inscribed  angle. 

Given  the  inscribed  /.BAG  (Fig.  3)  with  the  center  0 
outside  the  angle. 

To  prove  that  /BAG  is  measured  by  i arc  BC. 

Proof.  Let  the  pupil  supply  the  proof. 

259.  Note.  By  use  of  the  above  theorem,  if  the  number  of 
degrees  in  the  intercepted  arc  be  known,  the  number  of  degrees  in 
the  inscribed  angle  can  be  determined  immediately.  Thus,  if  the  arc 
BC  (Fig.  3)  contains  4S°,  the  angle  BAC  coutains  24°;  also,  if  it  be 
known  that  the  angle  BAC  contains,  say  27°,  the  arc  must  contain  54L 


260.  Cor.  1.  All  angles  inscribed  in  the  same  segment, 
or  in  equal  segments,  are  equal. 

Thus  A A,  A',  A"  (Fig.  4)  are  all  equal,  for  each  of 
them  is  measured  by  one -half  the  arc  BBC. 

261.  Cor.  2.  An  angle  inscribed  in  a semicircle  is  a 
right  angle,  for  it  is  measured  by  one-half  a semicircum- 
ference. 

Thus  FR  (Fig.  5)  is  a diameter  .*.  /FOR  is  a rt.  Z . 

262.  Cor.  3.  An  angle  inscribed  in  a segment  greater 
than  a semicircle  is  an  acute  angle;  an  angle  inscribed  in  a 
segment  less  than  a semicircle  is  an  obtuse  angle. 

Ex.  If,  in  Fig.  1,  p.  132,  Z A contains  23°.  how  many  degrees 
ere  there  in  arc  BC  t in  arc  AC  “ 


134 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XX.  Theorem 

263.  An  angle  formed  by  two  chords  intersecting  within 
a circumference  is  measured  by  half  the  sum  of  the  inter- 
cepted arcs. 


B 


Given  the  chords  AB  and  CD  in  the  O ADBC,  inter- 
secting within  the  circumference  at  the  point  P. 

To  prove  that  ZDPB  is  measured  by  £ (arc  DB  -f-  are 
AC). 

Proof.  Draw  the  chord  AD. 

Then,  in  A ADP, 

Z DPB  = Z A~f  Z D,  Art.  135. 

(an  ext.  Z of  a A is  equal  to  the  sum  of  the  two  opp.  int.  A ). 

But  Z ZL  is  measured  by  £ arc  DB,  Art.  258. 

(an  inscribed  Z is  measured  by  one-lialf  its  intercepted  arc). 

Also  ZD  is  measured  by  £ arc  A C.  (Why?) 

Hence  ZDPB  is  measured  by  4 (arc  DB  + arc  AC). 

Ax.  2. 

Q.  E.  D. 

Ex.  1.  In  the  above  figure,  if  are  DB  contains  64°  and  arc  AC  con- 
tains 38°,  how  many  degrees  in  AAPC  ? in  AAPD  ? 

Ex.  2.  If  arc  AD  = 52°  and  AAPD  = 124°,  find  arc  CB. 

Ex.  3.  If,  in  Fig.  1,  p.  132,  arc  AC  contains  112°,  how  many 
degrees  are  there  in  the  AAi 


MEASUREMENT  OF  ANGLES 


135 


Proposition  XXI.  Theorem 

264.  An  angle  formed  by  a tangent  and  a chord  drawn 
from  the  point  of  contact  is  measured  by  half  the  inter- 
cepted arc. 


Given  the  O PCD,  and  ZAPC  formed  by  the  tangent 
APB  and  the  chord  PC. 


To  prove  that  ZAPC  is  measured  by  J arc  PRC. 


Proof.  Let  the  chord  CD  be  drawn  |]  AB. 

Then 

Z APC  = Z PCD. 

( Whyf) 

But 

Z PCD  is  measured  by  J arc  PD. 

(Why  T) 

:.  Z APC  is  measured  by  J arc  PD. 

Ax.  8. 

But 

arc  PRC  = arc  DP, 

Art.  233. 

(two  ||  lines  intercept  equal  arcs  on  a circumference ) . 

Z APC  is  measured  by  £ arc  PRC. 

Ax.  8. 

Also 

ZBPC,  the  supplement  of  ZAPC,  is 

measured 

£ arc 

PDC, 

Ax.  3. 

(for  arc  PDC  is  the  conjugate  of  arc  PRC) . 

Q.  E.  D. 

Ex.  1.  In  the  above  figure,  if  arc  PRC  contains  124°,  how  many 
degrees  are  in  the  angle  APC  ? 

Ex.  2.  If  arc  CD  = 96°,  find  the  angles  on  the  figure. 


136 


BOOK  II.  PLANE  (TF.mrF.T-RV 


Proposition  XXII.  Theorem 

265.  An  angle  formed  by  two  secants , or  by  two  tangents_ 
or  by  a secant  and  a tangent  meeting  without  the  circumference, 
is  measured  by  half  the  difference  of  the  interested  arcs. 


I.  Given  the  O ACDB  (Fig.  1),  and  the  Z APB  formed 
by  the  two  secants  PA  and  PB , meeting  at  the  point  P with- 
out the  circumference. 

To  prove  that  Z P is  measured  by  £ (arc  AD — arc  CD) 


Proof.  Draw  the  chord  CB. 

Then  ZACB=ZP+ZB.  (Why?) 

:.  ZP=  ZACB— ZB.  Ax.  3. 

But  ZACB  is  measured  by  £ arc  AB.  Art.  258. 

Also  ZB  is  measured  by  £ arc  CD.  (Why?) 


.’.  ZP  is  measured  by  £ (arc  AB  — arc  CD) . Ax.  3. 

II.  Given  the  O ADB  (Fig.  2),  and  ZCPB  formed 
by  the  tangent  PC  and  the  secant  PB. 

To  prove  that  Z P is  measured  by  £ (arc  ABB  — arc  AD) . 
Proof.  Let  the  pupil  supply  the  proof. 


EXERCISES  ON  THE  CIRCLE 


137 


III.  Given  /.APB  (Fig.  3)  formed  by  the  tangents  PG 
and  PD. 

To  prove  that  /P  is  measured  by  £ (arc  AEB — ■ 
arc  APB) . 

Proof.  Let  the  pupil  supply  the  proof. 

266.  Note.  By  means  of  Props.  XVIII-XXII,  angles  rormed 
by  chords,  secants,  or  tangents,  or  combinations  of  these,  are  all 
reduced  to  central  angles  and  hence  may  readily  be  compared. 

Ex.  1.  If,  in  Fig.  1,  p.  136,  are  CD  = 34°  and  are  AB  — 108°, 
draw  AD  and  find  all  the  angles  of  the  figure. 

Ex.  2.  If,  in  Fig.  3,  p.  136,  angle  P— 80°,  find  arcs  AFB  and  AEB. 


EXERCISES.  CROUP  16 

Ex.  1.  What  new  methods  of  proving  two  lines  equal  are  afforded 
by  Book  II  ? 

Ex.  2.  What  new  methods  of  proving  two  angles  equal  ? of  prov- 
ing two  angles  supplementary  ? of  proving  an  angle  a right  angle  ? 


Ex.  3.  What  methods  of  proving  two  ares  equal  ? 

Ex.  4.  In  a quadrilateral  inscribed  in  a circle,  each  pair  of  opposite 
angles  is  supplementary. 

Ex.  5.  A chord  forms  equal  angles  with  the  tangents  at  its 
extremities. 


Ex.  6.  If  an  isosceles  triangle  be  inscribed  in  a circle,  the  tangent 
at  its  vertex  makes  equal  angles  with  two  of  its  sides  and  is  parallel 
to  the  third  side.  (Is  the  converse  of  this  theorem  true?) 


Ex.  7.  If  two  chords  in  a circle  intersect  within  the  circle  at  right 
angles,  the  sum  of  a pair  of  alternate  arcs  equals 
a semicircumference. 

Ex.  8.  Given  0 the  center  of  a circle  and 
AC  a tangent;  prove  Z.BAC=%  Z.  0. 

Ex.  9.  If  one  side  of  an  inscribed  quadri- 
lateral be  produced,  the  exterior  angle  so 
formed  equals  the  opposite  interior  angle  of  the 


138 


BOOK  II.  PLANE  GEOMETRY 


Ex.  10.  If  two  tangents  to  a circle  include  an  angle  of  60°  at 
their  point  of  intersection,  the  chord  joining  the  points  of  contact 
forms  with  the  tangents  an  equilateral  triangle. 

Ex.  1 1.  The  chord  AB  equals  the  chord  CD  in  a given  circle,  and 
the  chords,  if  produced,  intersect  at  the  point  P.  Prove  secant  PA  = 
secant  PC. 

Ex.  12.  A circle  is  circumscribed  about  the  triangle  ABC,  and  P is 
the  midpoint  of  the  are  AB.  Prove  that  the  angle  ABP  equals  one- 
half  the  angle  C. 

Ex.  13.  If  A,  B,  C,  D and  E be  points  taken  in  succession  on  the< 
circumference  of  a circle,  and  the  arcs  AB,  BC,  CD  and  DE  be  equal, 
prove  that  the  angles  ABC,  BCD  and  CDE  are  equal. 

Ex.  14.  An  inscribed  angle  formed  by  a diameter  and  a chord  has 
its  intercepted  arc  bisected  by  a radius  which  is 
parallel  to  the  chord. 

Ex.  15.  Two  secants,  PAB  and  PCD,  intersect 
the  circle  ABDC.  Prove  that  the  triangles  PBC 
and  PAD  are  mutually  equiangular. 

Ex.  16.  Given  AC  a tangent  and  AB\\CE; 
prove  A ACD  and  ABE  mutually  equiangular. 

Ex.  17.  Given  ABC  an  inscribed  triangle,  AE 
J_  BC,  and  CD  -L  AB  ; prove  arc  BD= arc  BE. 

Ex.  18.  If  the  diagonals  of  an  inscribed  quad- 
rilateral are  diameters,  the  quadrilateral  is  a 
rectangle. 

Ex.  19.  Tangents  through  the  vertices  of  an 
inscribed  rectangle  form  a rhombus. 

Ex.  20.  Two  circles  intersect  at  P 
and  Q.  PA  and  PB  are  diameters. 

Prove  that  QA  and  QB  form  a straight 
line. 

Ex.  21.  Two  equal  circles  intersect 
at  P and  Q,  through  P a line  is  drawn 
terminated  by  the  circumferences  at  A and  B.  Show  that  QA 
equals  QB. 

[gtJG.  A A and  B are  measured  by  what  arcs  T] 


C 


Exercises.  auxiliary  lines 


139 


267.  Use  of  auxiliary  lines.  In  demonstrating  theorems 
relating  to  the  circle,  it  is  often  helpful  to  draw  one  or 
more  of  the  following  auxiliary  lines:  A radius , a diame- 
ter, a chord,  a perpendicular  from  the  center  upon  a chord, 
an  arc,  a circumference,  etc. 

EXERCISES.  CROUP  17 

AUXILIARY  LINES 

Ex.  1.  Given  circle  0,  arc  AB= arc  BC, 

BM  L OA,  and  BN  L OC  ; prove  BM=BN. 

Ex.  2.  If  from  any  point  in  the  circumference 
of  a circle  a chord  and  a tangent  be  drawn,  the 
perpendiculars  drawn  to  them  from  the  midpoint 
of  the  arc  subtended  by  the  chord  are  equal. 

Ex.  3.  Given  0 the  center  of  a circle,  and  OD 
X chord  AC  ; prove  AAOD—  /.B. 

Ex.  4.  From  the  extremity  of  a diameter  chords 
are  drawn  making  equal  angles  with  the  diameter. 

Prove  that  these  chords  are  equal. 

[StTG.  Draw  _k  from  the  center  to  the  chords,  etc.] 

Is  the  converse  of  this  theorem  true  ? 

Ex.  5.  If  through  any  point  within  a circle  equal  chords  be  drawn, 
show  that  the  line  drawn  from  the  center  to  the  point  of  intersection 
of  the  chords  bisects  their  angle  of  intersection. 

Ex.  6.  Tangents  PA  and  PP  are  drawn  to  a circle  whose  center  is 
}0.  Prove  that  angle  P equals  twice  angle  OAB. 

Ex.  7.  If  in  a circle  two  equal  chords  intersect, 
the  segments  of  one  chord  equal  the  segments  of  the 

other  chord. 

Ex.  8.  A parallelogram  inscribed  in  a circle  is  a 

rectangle. 

[Sug.  Draw  the  diagonals  of  the  ZZ7 , use  Art.  218,  and  Ex,  21, 

P,  100.] 


A 


140 


BOOK  II.  PLANE  GEOMETRY 


Ex.  9.  If  a quadrilateral  be  circumscribed  about  a circle,  the  angles 
subtended  at  the  center  by  a pair  of  opposite  sides  are  supplementary 

[Sug.  Draw  radii  to  the  points  of  contact  and  show  that  there  are 
four  pairs  of  equal  A at  the  center.] 

Ex.  10.  If  an  equilateral  triangle  ABC  be  irfscribed  in  a circle  and 
any  point  P be  taken  in  the  are  AB,  show  that  PC=  PA-\-PB. 

[StJG.  On  PC  take  PM  equal  to  PA,  draw  AM  and  prove  A PAB 
and  MAC  equal.] 

Ex.  11.  Two  radii  perpendicular  to  each  other  are  produced  to 
intersect  a tangent,  and  from  the  points  of  intersection  other  tangents 
are  drawn  to  the  circle.  Prove  that  the  tangents  last  drawn  are 
parallel. 

[Sug.  Draw  radii  to  the  three  points  of  con- 
tact, etc.] 

Ex.  12.  A straight  line  intersects  two  concen- 
tric circles.  Show  that  the  segments  of  the  line 
intercepted  between  the  circumferences  are  equal 
(prove  AB=CD) . 

Ex.  13.  A common  tangent  is 
drawn  to  two  circles  which  are 
exterior  to  each  other.  Show  that 
the  chords  drawn  from  the  points 
of  tangency  to  the  points  where 
the  line  of  centers  cuts  the  cir- 
cumferences are  parallel. 

Ex.  14.  A circle  is  described  on  the  radius  of  another  circle  as  a 
diameter,  and  a chord  of  the  larger  circle  is  drawn  from  the  point  of 
contact  of  the  two  circles.  Prove  that  this  chord  is  bisected  by  the 
circumference  of  the  smaller  circle. 

[Sug.  If  the  chord  is  bisected,  a -L  from  the  center  of  the  larger  to 
the  chord  will  also  bisect  the  chord,  etc.] 

Ex.  15.  Two  circles  are  tangent  ex- 
ternally at  the  point  P.  Through  P any 
two  lines  APB  and  CPD  are  drawn  ter- 
minated by  the  circumferences.  Show 
that  the  chords  AC  and  BD  are  parallel. 

[Sug.  Draw  the  common  tangent  at 
P,  If  AC  and  BD  are  || , what  A must  be  equal  ? ] 


EXERCISES.  MAXIMA  AND  MINIMA 


141 


Ex.  16.  Two  circles  intersect  at  the  points  P and  Q.  Lines  APB 
and  CQD  are  drawn,  terminated  by  the  circumferences.  Show  that 
AC  and  BD  are  parallel. 

Ex.  17.  If  a square  be  described  on  the  hypotenuse  of  a right 
triangle,  a line  drawn  from  the  center  of  the  square  to  the  vertex  of 
the  right  angle  bisects  the  right  angle. 

[Sug.  Describe  a circumference  on  the  hypotenuse  of  the  right 
triangle  as  a diameter.] 

Ex.  18.  If  two  circles  are  tangent  externally  at  P,  and  a common 
tangent  touches  them  at  A and  B,  respectively,  the  angle  APB  is  & 
' right  angle. 

Ex.  19.  If  in  the  triangle  ABC  the  two  altitudes  BP)  and  AD  are 
drawn,  the  angle  ABD  equals  the  angle  AED. 

[Sug.  Describe  a semicireumferenee  on  AB  as  a diameter.] 

268.  Def.  A maximum  is  the  greatest  of  a class  of 
magnitudes  satisfying  certain  given  conditions,  and  a 
minimum  is  the  least  (see  Arts.  470,  471).  For  instance,  of 
the  chords  in  a given  circle,  which  is  the  maximum  ? 

EXERCISES.  CROUP  18 

MAXIMA  AND  MINIMA 

Ex.  1 . Of  the  chords  drawn  through  a given  point  within  a circle, 
determine  which  is  the  greatest,  and  also  which  is  the  least. 

Ex.  2.  Find  the  shortest  line,  and 
also  the  longest  line,  that  can  be  drawn 
from  a given  external  point  to  the  cir- 
cumference of  a circle. 

[Sug.  0 being  the  center,  prove 
in  A PCO,  PA  < PC  ; by  A PDO, 

PB  > J1D.] 

Ex.  3.  Find  the  shortest  line,  and  als 
est  line,  that  can  be  drawn  to  the  eir 
of  a circle  from  a point  within  the  circ 

[Sug.  O being  the  center,  prove, 

A OPC,  PB  < PC,  etc.] 


142 


BOOK  II.  PLANE  GEOMETRY 


~/?- 


-“\P. 


0 

1 O’ 

Ex.  4.  If  two  circles  intersect,  show 
that,  of  lines  drawn  through  a point  of 
intersection  and  terminated  by  the  cir- 
cumferences, that  line  is  a maximum 
which  is  parallel  to  the 
. A line  of  centers. 

P [Sug.  Prove  ES  < 00’  CPD  < APB.] 

Ex.  5.  Given  AJS  L OB  in  circle  0 ; prove 
Z OAB  the  maximum  of  all  A.  having  their  vertices 
on  the  circumference  and  their  sides  passing  through 
0 and  P respectively. 

[Sug.  Draw  a circle  on  OA  as  a diameter.] 


EXERCISES.  CROUP  19 

DEMONSTRATIONS  BY  INDIRECT  METHODS 

Prove  the  following  by  an  indirect  method  (see  Art.  195) : 

Ex.  1.  A segment  of  a circle  which  contains  a right  angle  is  a 
semicircle. 

Ex.  2.  If  a rectangle  be  inscribed  in  a circle,  its  diagonals  are 
diameters. 

Ex.  3.  Prove  the  second  part  of  Prop.  VI  by  an  indirect  method. 

Ex.  4.  Prove  Prop.  X by  an  indirect  method. 

Ex.  5.  A straight  line  connecting  the  midpoint  of  a chord  and  the 
midpoint  of  the  arc  subtended  by  the  chord  is  perpendicular  to  the 
chord  (use  the  method  of  coincidence). 

Ex.  6.  A line  joining  the  midpoints  of  two  parallel  chords  passes 
through  the  center. 

[Sug.  Draw  a 1 to  each  chord  from  the  center  and  show  that 
these  A are  in  the  same  line,  etc.] 

Ex.  7.  If  the  opposite  angles  of  a quadrilateral  are  supplementary, 
a circle  can  be  circumscribed  about  the  quadrilateral. 

[Sug.  Pass  a circumference  through  three  vertices  of  the  quadri- 
lateral; if  it  does  not  pass  through  the  remaining  vertex,  etc.] 

Ex.  8.  Prove  the  converse  sf  Prop.  XII. 


EXERCISES.  CONSTANTS  AND  LOCI 


143 


269.  A geometrical  constant  is  a geometrical  magni- 
tude which  varies  in  some  respect,  as  in  position,  but 
remains  constant  in  size.  Thus  the  angles  inscribed  in  a 
given  semicircle  vary  in  position  but  are  all  of  the  same 
size;  viz.,  a right  angle  (see  Art.  261). 

EXERCISES.  CROUP  20 

DETERMINATION  OF  CONSTANTS  AND  LOCI 

Ex.  1.  AB  and  AC  are  tangents  to  a circle.  Pis  any  point  on  the 
circumference  outside  the  triangle  ABC.  As  P moves,  prove  that  the 
sum  of  the  LA  and  LBPC  is  constant. 

Ex.  2.  In  Ex.  8,  p.  123,  if  AB  and  BQ  be  produced  to  meet  at  T, 
show  that  the  perimeter  of  the  triangle  TBQ  equals  the  sum  of  TA 
and  TB,  and  hence  that  the  perimeter  of  triangle  TBQ  is  constant,  no 
matter  how  P may  vary  in  position  between  A and  B. 

Ex.  3.  Show  on  the  same  figure  that,  if  O is  the  center,  LBOQ  is 
constant  as  P varies  in  position. 

Ex.  4.  Two  circles  intersect  in  the  points 
A and  B.  From  any  point  P on  one  circum- 
ference lines  PAC  and  PBD  are  drawn,  ter- 
minated by  the  other  circumference.  Show 
that  the  chord  CD  is  constant. 

[Sug.  Draw  BC  and  prove  Z CBD  a 
eonstanfr.] 

Ex.  5.  Given  OA  L OB,  and  CD  a line  of 
1 given  length  moving  so  that  D is  always  in  OA 
and  C in  OB  and  P the  midpoint  of  CD  ; prove 
that  OP  is  constant  in  length  (see  Ex.  G,  p.  94). 

The  determination  of  loci  is  often 
facilitated  by  showing  that  some  given 
magnitude  is  a constant. 

Ex.  6.  Find  the  locus  of  a point  moving  so  that  it  is  at  a given 
distance  a from  a given  circumference  whose  radius  is  r. 


144 


BOOK  II.  PLANE  GEOMETRY 


Ex.  7.  Find  the  locus  of  the  midpoints  of  the  radii  of  a given 
circle. 

Ex.  8.  Find  the  locus  of  the  midpoints  of  all  chords  of  a given 
length  drawn  in  a given  circle. 

Ex.  9.  Find  the  locus  of  the  vertices  of  all  right  triangles  having 
a given  hypotenuse. 

Ex.  10.  Find  the  locus  of  the  midpoints  of  all  the  chords  drawn 
from  a given  point  on  a given  circumference. 

[Sug.  Draw  a line  from  the  center  to  the  given  point  and  perpen- 
diculars from  the  center  upon  the 
chords.] 

Ex.  1-1.  In  Ex.  5,  p.  143,  find 
the  locus  of  P. 

Ex.  12.  QP  is  a line  of  given 
length  and  moves  so  that  Q is 
always  in  a given  circumference,  and  QP  is  always  parallel  to  a fixed 
line.  Find  the  locus  of  P. 


EXERCISES.  CROUP  21 

THEOREMS  PROVED  BY  VARIOUS  METHODS 

Ex.  1.  The  line  which  bisects  the  angle  formed  by  a tangent  and 
a chord  bisects  the  intercepted  are  also. 

Ex.  2.  An  inscribed  trapezoid  is  isosceles. 

Ex.  3.  Given  TA  and  TB  tangents,  arc  AB  = 

80°,  arc  BD= 95°,  and  arc  DC=  150°;  find  all  the 
angles  of  the  figure. 

Ex.  4.  If  two  tangents  to  a circle  are  parallel, 
the  line  joining  their  points  of  contact  is  a diameter. 

[Sug.  Draw  radii  to  the  points  of  contact  and 
use  an  indirect  method  of  proof.] 


Ex.  5.  A rectangle  circumscribed  about  a circle  is  a square. 
[Sug.  Use  the  preceding  theorem.] 


MISCELLANEOUS  EXERCISES 


145 


Ex.  6.  Given  0 the  center  of  a circle, 
and  BP  = the  radius;  prove  AAOC=  3ZP. 

Ex.  7.  Find  the  angle  formed  by  the  q 
side  of  an  inscribed  square  and  the  tangent 
through  the  vertex  of  the  square. 

Ex.  8.  From  an  external  point  a secant  is  drawn  through  th^, 
center  of  a circle,  and  also  two  other  secants  making  equal  angles 
with  the  first  secant.  Show  that  the  secants  last  drawn  are  equal. 

Ex.  9.  ABC  is  a triangle  and  on  the  side  AB  the  point  P is 
taken  and  on  BC  the  point  Q,  so  that  angle  BPQ  equals  angle  C. 
Show  that  a circle  may  be  circumscribed  about  the  quadrilateral 
APQC. 

Ex.  10.  Two  circles  are  tangent  to  each  other  externally,  and  a 
line  is  drawn  through  the  point  of  contact  terminated  by  the  circum- 
ferences. Show  that  the  radii  from  the  extremities  of  this  line  are 
parallel. 

Ex.  11.  If  a circumference  be  described  on  the  leg  of  an  isosceles 
triangle  as  diameter,  the  circumference  will  bisect  the  base  of  the 
triangle. 

Ex.  12.  The  chord  of  an  arc  is  parallel  to  the  tangent  at  the 
midpoint  of  the  arc. 

Ex.  13.  If  a triangle  be  inscribed  in  a circle,  the  sum  of  the 
angles  inscribed  in  the  segments  exterior  to  the  triangle  is  four  right 
angles. 

Ex.  14.  Find  the  corresponding  theorem  for  an  inscribed 
quadrilateral. 

Ex.  15.  Find  the  locus  of  the  centers  of  all  circles  passing  through 
two  given  points. 

Ex.  16.  If  two  unequal  chords  intersect  in  a circle,  the  greater 
chord  makes  the  less  angle  with  the  diameter  through  the  point  of 
intersection  of  the  chords. 

State  also  the  converse  of  this  theorem.  Is  the  converse  true  ? 

Ex.  17.  The  sum  of  the  legs  of  a right  triangle  equals  the  sum  of 
the  hypotenuse  and  the  diameter  of  the  inscribed  circle. 

J 


146 


BOOK  II.  PLANE  GEOMETRY 


Ex.  18.  The  sides  AB,  BC  and  AC  of  a triangle  touch  the  in- 
scribed circle  at  the  points  P,  Q and  M.  Show  that  angle  PQR  and 
one -half  angle  A are  complementary. 

[Sug.  Draw  radii  from  the  center  O to  P and  B.  Then  ZPQB  = 
Z POA,  etc.] 

Ex.  19.  From  the  point  in  which  the  bisector  of  an  inscribed 
angle  meets  the  circumference,  a chord  is  drawn  parallel  to  one  side 
of  the  angle.  Show  that  this  chord  equals 
the  other  side  of  the  angle. 

Ex.  20.  Given  AB  a diameter,  AP— 
the  radius,  AD  and  PC  tangents ; prove 
A CED  equilateral. 

[Sug.  Draw  OC  and  CA;  then  in  rt. 

A OCP,  Cb4  = radius,  ZP= 30°,  etc.] 

Ex.  21.  Perpendiculars  are  drawn 
from  the  extremities  of  a diameter  upon 
a tangent.  Show  that  the  points  in  which  the  perpendiculars  intersect 
the  tangent  are  equidistant  from  the  center. 


CONSTRUCTION  PROBLEMS 

270.  Postulates.  As  stated  in  Art.  49,  a postulate  in 
geometry  is  a construction  of  a geometric  figure  admitted 
as  possible. 

The  postulates  used  in  geometry,  are  as  follows  (see 
Art,  50): 

1.  Through  any  two  points  a straight  line  may  he  drawn. 

2.  A straight  line  may  he  extended  indefinitely , or  it  may 
he  limited  at  any  given  point. 

3.  A circumference  may  he  described  about  any  given 
point  as  a center  and  with  any  given  radius. 


CONSTRUCTION  PROBLEMS 


147 


271.  The  meaning  of  the  postulates  is  that  only  two 
drawing  instruments  are  to  be  used  in  making  geometri- 
cal constructions;  viz.,  the  straiglit-edge  ruler  and  the 
compasses. 

With  these  two  simplest  drawing  instruments  it  is 
desirable  to  be  able  to  construct  as  many  geometrical 
figures  as  possible. 

272.  Form  of  solution  of  a problem.  The  statement  of 
a problem  and  of  its  solution  consists  of  certain  distinct 
parts  which  it  is  important  to  keep  in  mind.  These  are 

1.  The  general  enunciation 

2.  The  particular  enunciation. 

(1)  Given,  etc. 

(2)  To  construct,  etc.  (or  some  other  construction 

phrase,  as  "to  draw,”  "to  bisect,”  etc.). 

3.  The  construction. 

4.  The  assertion. 

5.  The  proof  of  the  assertion. 

6.  The  conclusion  (indicated  by  Q.  E.  f.,  quod  erat 
faciendum,  "which  was  to  be  done”). 

7.  The  discussion  of  special  or  limiting  cases,  if  such 
cases  occur  in  the  given  problem. 


In  the  figures  drawn  in  connection  with  problems,  the  given  lines 
are  drawn  as  heavy  lines,  the  lines  required  as  light  lines , and  the 
auxiliary  lines  as  dotted  line*. 


148 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XXIII.  Problem 

273.  From  a given  point  without  a given  line  to  draw 
a perpendicular  to  the  line. 

P 

„ Q. 

s / 


P 

A B 

Given  the  line  AB  and  the  point  P outside  AB. 

To  construct  a perpendicular  from  the  point  P to  the 
line  AB. 

Construction.  With  P as  a center  and  with  any  conve- 
nient radius,  describe  an  arc  intersecting  AB  in  two  points  as 
at  C and  D.  Post.  3. 

From  C and  D as  centers  and  with  convenient  equal 
radii,  greater  than  4 CD,  describe  arcs  intersecting  at  Q. 

Post.  3. 

Draw  the  line  PQ  and  produce  it  to  meet  AB  at  JR. 

Posts.  1,  2. 

[Assertion].  Then  PR  is  the  J_  required. 

1 Proof.  P is  equidistant  from  the  points  C and  D.  Constr. 

Also  Q is  equidistant  from  the  points  C and  D.  Constr 

.*.  PR  _L  CD.  Art.  113. 

(two  points,  each  equidistan  t from  the  extremities  of  a line,  determine  the 
X bisector  of  the  line) . 


Q.  E.  F. 


CONSTRUCTION  PROBLEMS 


149 


Proposition  XXIV.  Problem 

274.  At  a given  point  in  a given  line  to  erect  a per- 
pendicular to  that  line. 


Pig.  1 Fig.  2 


Given  the  point  P in  the  line  AB 

To  construct  a perpendicular  to  the  line  AB  at  the  point  P. 

Method  I.  Construction.  From  P (Fig.  1)  as  a center 
with  a convenient  radius,  describe  an  arc  cutting  off  the 
equal  segments  PC  and  PD  on  the  line  AB.  Post.  3. 

From  C and  D as  centers  and  with  equal  radii,  greater 
than  PD,  describe  arcs  intersecting  at  R.  Post.  3. 

Draw  PR.  Post.  1. 

[Assertion] . Then  PR  is  the  _L  required. 

Proof.  Let  the  pupil  supply  the  proof. 

Method  II.  Construction.  Take  any  point  0 (Fig.  2) 
without  the  line  AB,  and  with  OP  as  a radius,  describe  a 
circumference  intersecting  the  line  AB  at  C.  Post.  3. 

Draw  CO,  and  produce  CO  to  meet  the  circumference 
at  R.  Draw  RP.  Posts,  i and  2. 

[Assertion] . Then  RP  is  the  _L  required . 

Proof.  Let  the  pupil  supply  the  proof. 

Discussion,  When  is  Method  II  preferable  ? 


150 


275. 


'XD 

Given  the  line  AB. 

To  bisect  line  AB. 

Construction.  With  A and  B as  centers  and  with  equal 
radii,  greater  than  £■  AB,  describe  arcs  intersecting  in 
C and  D.  Post.  3. 

Draw  the  line  CD  intersecting  AB  in  F.  Post.  1. 

Then  AB  is  bisected  at  the  point  F. 

Proof.  Let  the  pupil  supply  the  proof. 

Q.  E F. 


BOOK  II.  PLANE  GEOMETB* 

Proposition  XXV.  Problem 
To  bisect  a given  line. 

y? 


Proposition  XXVI.  Problem 
276.  To  bisect  a given  arc. 


D 


Given  the  arc  AB. 
To  bisect  arc  AB. 


CONSTRUCTION  PROBLEMS 


151 


Construction.  With  A and  B as  centers,  and  with  con- 
venient equal  radii,  describe  arcs  intersecting  at  G and  D. 

Post.  3. 

Draw  the  line  CD  intersecting  the  arc  AB  at  F.  Post.  l. 
Then  arc  AB  is  bisected  at  the  point  F. 

Proof.  Draw  the  chord  AB. 

Then  CD  J_  chord  AB  at  its  middle  point.  (Why?) 

.-.  CD  bisects  the  arc  AB , Art.  223. 

(the  i.  bisector  of  a chord  passes  through  the  center  and  bisects  the  arcs 
subtended  by  the  chord). 


Q.  E.  F 


Proposition  XXVII.  Problem 
277.  To  bisect  a given  angle. 


Given  angle  AOB. 

To  bisect  angle  AOB. 

Construction.  With  0 as  a center  and  with  any  conve- 
nient radius,  describe  an  arc  intersecting  OA  at  C and 
OB  at  D.  Post.  3. 

With  C and  D as  centers  and  with  convenient  equal 
radii,  describe  arcs  intersecting  at  F. 

Draw  OF. 

Then  Z AOB  is  bisected  by  line  OF. 

Proof.  Let  the  pupil  supply  the  proof. 

Sx.  Construct  an  Z of  45°. 


Post.  3. 

Post.  1. 

Q.  8.  F. 


152 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XXVIII.  Problem 


278.  At  a given  point  in  a given  straight  line  to  con- 
struct an  angle  equal  to  a given  angle. 


Given  the  Zl,  and  the  point  P in  the  line  BC. 

To  construct  at  the  point  P an  angle  equal  to  A A,  and 
having  PC  for" one  of  its  sides. 

Construction.  With  A as  a center  and  with  any  conve- 
nient radius,  describe  an  arc  meeting  the  sides  of  A A 

at  B and  E.  Post.  3. 

Draw  the  chord  BE.  Post.  1. 

With  P as  a center  and  with  a radius  equal  to  AB, 
describe  an  arc  cutting  PC  at  F.  Post.  3. 

From  F as  a center  and  with  a radius  equal  to  the  chord 
BE,  describe  an  arc  intersecting  the  arc  HE  at  G.  Post.  3. 

Draw  PG.  Post.  1. 

Then  A GPF  is  the  angle  required. 

Proof.  Let  the  pupil  draw  the  chord  EG  and  complete 
the  proof. 

Q.  E.  F. 


Ex.  1.  On  a given  line  construct  a square. 
Ex.  2.  Construct  an  angle  of  30°. 

Ex.  3.  Hence  construct  an  angle  of  15°. 


CONSTRUCTION  PROBLEMS 


153 


Proposition  XXIX.  Problem 

279.  Through  a given  point  without  a given  straight 
line  to  draw  a line  parallel  to  a given  line. 

D 


7\ 

/ \ 


7-n 


C 

Given  any  point  P without  the  line  AB. 

To  construct  a line  through  P ||  AB. 

Construction.  Through  P draw  any  convenient  line  CD 

meeting  AB  in  C.  Post.  l. 

At  P in  the  line  CD  construct  /.DBF  equal  to  ZPCB. 

Art.  278. 

Then  EF  is  the  line  required. 

Proof.  Let  the  pupil  supply  the  proof. 


Ex.  1.  Through  a given  point  draw  a line  parallel  to  a given  line 
by  constructing  a parallelogram  of  which  the  given  point  is  one 
vertex. 

Ex.  2.  On  a given  line  as  a diameter,  construct  a circle. 

Ex.  3.  Construct  an  arc  of  60°  having  a radius  of  1 in. 

Ex.  4.  On  the  figure,  p.  148,  if  the  point  Q be  constructed  below 
the  line  AB,  will  the  perpendicular  required  be  likely  to  be  more 
accurately,  or  less  accurately  constructed  ? 

Ex.  5.  From  a given  point  on  a given  circumference,  how  many 
equal  chords  can  be  drawn  ? 

Ex.  6.  Through  a given  point  within  a given  circumference,  how 
many  equal  chords  can  be  drawn? 

Ex.  7.  From  a given  point  external  to  a given  circle,  how  many 
equal  secants  can  be  drawn  ? 


154 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XXX.  Problem 

280.  To  divide  a given  straight  line  into  any  required 
number  of  equal  parts. 

AFC,  B 


Given  the  line  AB. 

To  divide  AB  into  a required  number  (as  three)  equal 
parts. 

Construction.  From  A draw  the  line  AP  making  a con- 
venient angle  with  AB.  Post.  l. 

Take  AC  any  line  of  convenient  length  and  apply  it  to 
AP  a number  of  times  equal  to  the  number  of  parts  into 
which  AB  is  to  be  divided.  Post.  2. 

From  E,  the  end  of  the  measure  when  last  applied  to 
AP,  draw  EB.  Post.  l. 

Through  the  other  points  of  division  on  AP,  viz.,  C 

and  D,  draw  lines  1 1 EB  and  meeting  AB  at  F and  O. 

Art,  279 

Then  AB  is  divided  into  the  required  number  of  parts 
at  F and  O. 

Proof.  AC  = CD  = DE.  Constr. 

AF  = FO  = GB,  Art.  176. 

(»/  three  or  more  parallels  intercept  equal  parts'  on  one  transversal,  they 
intercept  equal  parts  on  every  transversal) . 


Ex.  1.  Construct  a right  triangle  whose  legs  are  1 in.  and  lj  in. 
Ex.  2.  Construct  a rectangle  whose  base  is  2 in.  and  altitude  1 in. 


CONSTEUCTION  PROBLEMS 


155 


Proposition  XXXI.  Problem 


281.  To  construct  a triangle,  given  two  sides  and  the 
included  angle. 


■B 


Given  m and  n two  sides  of  a triangle  and  P the  angle 
included  by  them. 

To  construct  the  triangle. 


Construction.  At  the  point  A in  the  line  AB  construct 
LA  equal  to  the  given  Z P.  Art.  278. 


On  the  line  AB  lay  off  AO  equal  to  m. 
On  the  line  AF  lay  off  AD  equal  to  n. 
Draw  DC. 


Post.  2. 
Post.  2. 
Post.  1. 


The  A ADC  is  the  triangle  required. 


Q.  E.  F. 


Ex.  1.  Construct  a triangle  in  which  two  of  the  sides  are  1 in.  and 
2»ln.,  and  the  included  angle  is  135°. 

Ex.  2.  Construct  an  isosceles  triangle,  in  which  the  base  shall  be 
li  in.  and  the  altitude  2 in. 

Ex.  3.  Construct  the  complement  of  a given  acute  angle. 

Ex.  4.  Construct  the  supplement  of  a given  angle. 

Ex.  6.  How  is  the  figure  on  p.  154  constructed  with  the  fewest 
adjustments  of  the  compasses  f 

Ex.  §.  Draw  a line  (segment)  and  mark  off  three-fifths  of  it 


156 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XXXII.  Problem 


282.  To  construct  a triangle,  given  two  angles  and  the 
included  side. 


c 


m 

Given  the  A P and  Q and  the  included  side  m. 


To  construct  the  triangle. 


Construction.  Take  any  line  AD  and  on  it  mark  AB 
equal  to  m.  Post.  2. 

At  A construct  an  angle  equal  to  ZP.  Art.  278. 

At  B construct  an  angle  equal  to  Z Q.  Art.  278. 


Produce  the  sides  of  the  A A and  B to  meet  at  C. 

Post.  2. 

Then  A ACB  is  the  triangle  required. 

Q.  E.  F. 


Discussion.  Is  it  possible  to  construct  the  triangle  if  the 
sum  of  the  given  angles  is  two  right  angles  ? Why  ? Is  it 
possible  if  this  sum  is  greater  than  two  right  angles  ? Why  ? 


Ex.  1.  Construct  a triangle  in  which  two  of  the  angles  are  30°  and 
45°,  and  the  included  side  is  It  in. 

Ex.  2.  Construct  the  complement  of  half  a given  angle. 

Ex.  3.  Construct  an  angle  of  120° ; of  150°;  of  135°. 

Ex.  4.  Trisect  a given  right  angle. 


CONSTRUCTION  PROBLEMS 


157 


Proposition  XXXIII.  Problem 
283.  To  construct  a triangle,  given  the  three  sides, 

m 

n 

P 


Given  m , n andp,  the  three  sides  of  a triangle. 

To  construct  the  triangle. 

Construction.  Take  the  line  AB  equal  to  m.  Post.  2. 

With  A as  a center  and  with  a radius  equal  to  n describe 
an  arc,  and  with  B as  a center  and  with  a radius  equal  to 
p describe  another  arc.  Post.  3. 

Let  the  two  arcs  intersect  at  the  point  C. 

Draw  GA  and  CB.  Post.  i. 

Then  A ABC  is  the  triangle  required. 

Discussion.  Is  it  possible  to  construct  the  triangle  if  one 
of  the  sides  is  greater  than  the  sum  of  the  other  two  sides  ? 
Why  t 

What  kind  of  a figure  is  obtained  if  one  side  equals  the 
sum  of  the  other  two  sides  ? 


Ex.  1.  Construct  an  angle  of  221°. 

Ex.  2 Divide  a given  circumference  into  four  quadrants. 

Ex.  3.  Construct  the  figure  on  page  82,  using  the  concurrence  of 
the  three  altitudes  as  a teat  of  the  accuracy  of  the  work. 


158 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XXXIV.  Problem 

284.  To  construct  a triangle,  given  two  sides  and  an 
angle  opposite  one  of  them. 

Given  m and  n two  sides  of  a A and  ZP  opposite  n. 

To  construct  the  triangle. 

Construction.  Several  cases  occur,  according  to  the  rela- 
tive size  of  the  given  sides  and  the  size  of  the  given  angle. 

Case  I.  When  n > m ( and  ZP  is  acute). 


At  the  point  A construct  /LEAD  equal  to  ZP.  Art.  278. 

On  AE  take  AB  equal  to  m.  Post.  2. 

With  B as  a center  and  with  a radius  equal  to  n,  describe 
an  arc  intersecting  AD  at  C and  C' . Post.  3. 

Draw  BC  and  BC'.  Post.  1. 

Two  A,  ABC  and  ABC',  are  obtained,  containing  the 
sides  m and  n;  but  only  one  of  them,  A ABC,  contains 
the  ZP. 

/.  A ABC  is  the  triangle  required. 

Case  II.  When  n — m ( and  ZP  is  acute) . 

Make  the  same  construction  as  in  the  preceding  case. 

The  arc  drawn  intersects  the  line  AD  in  the  points  A 
and  C. 

Hence  the  isosceles  A ABC  is  the  triangle  required. 


CONSTRUCTION  PROBLEMS 


159 


Case  III.  When  n < m ( and  /.Pis  acute). 


m 


Make  the  construction  in  the  same  way  as  in  Case  I. 

Two  A,  ABC  and  ABC',  are  obtained,  each  of  which 
contains  the  sides  m and  n and  an  angle  equal  to  Z P oppo- 
site the  side  n. 

A ABC  and  ABC'  are  the  triangles  required. 

Discussion.  In  Case  I,  if  Z P is  a right  Z , let  the  pupil 
construct  the  figure  and  show  that  there  are  two  A answer- 
ing the  given  conditions.  If  ZP  is  an  obtuse  angle,  let 
him  construct  the  figure  and  show  that  there  is  but  one 
answer. 

In  Case  II,  if  ZP  is  right,  or  obtuse,  what  results  are 
obtained  ? 

In  Case  III,  if  ZP  is  acute  and  n = the  JL  from  B to 
AD,  how  many  answers  are  there?  also,  if  n < this  J_, 
how  many  ? 

If  ZP  is  right,  or  obtuse,  what  result  is  obtained? 


Ex.  1.  Construct  a triangle  in  which  two  of  the  sides  are  1 in.  and 
H in.,  and  the  angle  opposite  the  latter  side  is  45°. 

Ex.  2.  Construct  a triangle  in  which  two  of  the  sides  are  11  in. 
and  1 in.,  and  the  angle  opposite  the  latter  side  is  45°. 

Ex.  3.  Construct  a triangle  in  which  two  of  the  sides  are  li  in. 
and  f in.,  and  the  angle  opposite  the  latter  side  is  30°. 

Ex.  4.  Construct  the  figure  of  page  80,  using  the  concurrence  of 
the  three  bisectors  as  a test  of  the  accuracy  of  the  work. 


160 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XXXV.  Problem 


285.  To  construct  a parallelogram , given  two  sides  and 
the  included  angle. 


Given  m and  n two  sides,  and  P the  included  Z of  a ZZ7 . 
To  construct  the  parallelogram. 

Construction.  Take  line  AB  equal  to  m.  Post.  2. 

At  the  point  A construct  Z CAB  equal  to  ZP.  Art.  278 
On  the  side  AC  lay  oft  AB  equal  to  n. 

From  B as  a center  and  with  a radius  equal  to  m,  and 
from  B as  a center  with  a radius  equal  to  n,  describe  arcs 
intersecting  in  E.  Post.  3. 

Draw  EB  and  EB.  Post.  1. 

Then  ABEB  is  the  parallelogram  required. 

Proof.  Let  the  pupil  supply  the  proof. 


Proposition  XXXVI.  Problem 
286.  To  circumscribe  a circle  about  a given  iriangU 


Given  the  A ABC. 

To  construct  a circumscribed  O about  ABO. 


CONSTRUCTION  PROBLEMS 


161 


Construction.  Erect  J*  DF  and  FG  at  the  midpoints  of 
the  sides  AG  and  AB,  respectively.  Art.  274. 

From  0,  the  point  of  intersection  of  these  _k , with  a 
radius  equal  to  OA,  describe  the  O ABC.  Post.  3,  Art.  197. 
Then  O ABC  is  the  circle  required. 

Proof.  Let  the  pupil  supply  the  proof. 

Q.  E.  F. 

Proposition  XXXVII.  Problem 
287.  To  inscribe  a circle  in  a given  triangle. 


BT 


Given  the  A ABC. 

To  construct  an  inscribed  circle  in  the  triangle. 
Construction.  Draw  the  line  AD  bisecting  ZB  AC,  and 
CE  bisecting  ZBCA.  Art.  277. 

From  0,  the  intersection  of  AD  and  CE,  draw  the  line 
OP  JL  AC.  Art.  273. 

From  0 as  a center  with  a radius  OP,  describe  a O . 

Post,  3. 

Then  circle  0 is  the  circle  required. 

Proof.  Let  the  pupil  supply  the  proof. 

Q.  E.  F. 

Ex.  1.  Find  the  center  of  a given  circumference. 

Ex.  2.  Construct  the  figure  on  page  81,  using  the  concurrence  of 
the  three  perpendicular  bisectors  as  a test  of  the  accuracy  of  the 

work. 


162 


BOOK  II.  PLANE  GEOMETBY 


Proposition  XXXVIII.  Problem 

288,  Through  a given  point  on  the  circumference  to  draw 
a tangent  to  a circle. 


Given  any  point  P on  the  circumference  of  the  circle  0. 

To  construct  a tangent  to  the  circle  at  the  point  P. 

Construction.  Draw  the  radius  OP. 

At  the  point  P construct  a line  AB  J_  OP.  Art.  274. 

Then  AB  is  the  tangent  required. 

Proof.  Let  the  pupil  supply  the  proof. 

289.  An  escribed  circle  is  a circle 
tangent  to  one  side  of  a triangle  and 
to  the  other  two  sides  produced. 

Thus  the  circle  0 is  an  escribed 
circle  of  the  A ABC.  A center  of  an 
escribed  circle,  as  0,  is  called  an  ex- 
center of  the  triangle. 

Ex.  1.  Draw  a triangle  and  all  of  its  escribed  circles. 

Ex.  2.  Construct  the  figure  on  page  88,  using  the  concurrence  of 
the  three  medians  as  a test  of  the  accuracy  of  the  work. 


CONSTEUCTION  PROBLEMS 


163 


Proposition  XXXIX.  Problem 

290.  F*-om  a given  point  without  a circle  to  draw  a tan 
gent  to  the  circle. 


Given  P any  point  without  the  circle  0. 

To  construct  a line  through  P tangent  to  the  circle  0. 

Construction.  Draw  the  line  PO.  Post.  1. 

Bisect  the  line  PO  at  M.  Art.  275. 

From  M as  a center  with  MP  as  a radius,  describe  a cir- 
cumference intersecting  the  given  circumference  at  A 
and  B.  Post.  3. 

Draw  PA  and  PB. 

Then  PA  and  PB  are  the  tangents  required. 

Proof.  LPAO  is  inscribed  in  a semicircle. 

.*.  LPAO  is  a right  angle. 

.*.  PA  is  tangent  to  the  circle  0. 

In  like  manner  PB  is  tangent  to  the  circle  0. 


Post.  1. 

Constr. 

(Why?) 

(Why?) 

9.  e.  r. 


164 


BOOK  II.  PLANE  GEOMETRY 


Proposition  XL.  Problem 

291.  Upon  a given  straight  line  to  describe  a segment 
which  shall  contain  a given  inscribed  angle. 


Given  the  straight  line  AB  and  the  ZP. 

To  construct  on  the  line  AB  a segment  of  a circle  such 
that  an\'  angle  inscribed  in  the  segment  shall  equal  ZP. 

Construction.  At  the  point  B in  the  line  AB  construct 
Z ABC  equal  to  ZP.  Art.  278. 

Construct  BE  the  J_  bisector  of  line  AB.  Art.  274. 

At  B construct  BO  _L  BC,  and  intersecting  BE  at  0. 

Arts.  274,  122. 

From  0 as  a center  with  OB  as  a radius,  describe  the 
circle  AMB  Post.  3. 

Then  AMB  is  the  required  segment. 

Proof.  Let  AQB  be  any  Z inscribed  in  the  segment 
AQB. 


Then 

ZAQB  is  measured  by  J arc  AFB. 

Art.  258. 

But 

BC  is  tangent  to  the  circle  AMB. 

(Why?) 

Z ABC  is  measured  by  4 arc  AFB. 

(Why?) 

ZQ=  Z ABC,  or  Z P. 

(Why?) 

any  Z inscribed  in  segment  AMB  = ZP. 

Q.  E,  F. 


CONSTRUCTION'  PROBLEMS; 


165 


Proposition  XLI.  Theorem. 


292.  To  find  the  common  unit  of  measure  of  ho o commen- 
surable straight  lines,  and  hence  the  ratio  of  the  lines . 


A I — — - i — 

I 

Cl 1 1 t-H  D 

L 


IC 

—I — I— I — HS 

P 


Given  two  lines  AB  and  CD,  of  wliicli  CD  is  the  shorter 

To  construct  a common  unit  of  measure  of  AB  and  CD, 
and  hence  obtain  the  ratio  of  AB  and  CD. 

Construction.  Apply  CD  to  AB  as  many  times  as 
possible;  Post.  2. 

Say  twice,  with  a remainder  KB. 

Then  apply  AS  to  CD  as  many  times  as  possible ; Post.  2. 
Say  three  times,  with  a remainder  LD. 

Apply  LD  to  KB  as  many  times  as  possible;  Post.  2. 
Say  three  times,  with  a remainder  PB. 


Apply  PB  to  LD  as  many  times  as  possible;  Post.  2. 
Say  it  is  contained  in  LD  exactly  twice. 

Then  PB  is  the  common  unit  of  measure  of  AB  and  CD . 

Proof.  LD  = 2PB. 


Hence 


KB  — KP  + PB  = 3LD  + PB  = 7PB.  Axs.  6,  8. 
CD  = CL  + LD  = 3 KB  + LD  = 23  PB. 

AB  = AKJrKB  = 2CD  + KB= 53PB. 

AB  = 53  PB  = 53 
CD  23  PB  23 


Q.  E.  F. 


166 


BOOK  II.  PLANE  GEOMETRY 


EXERCISES  IN  CONSTRUCTION  PROBLEMS 

293.  Analysis  of  problems.  The  method  of  analysis 
(see  Art.  196)  is  of  especial  value  in  the  solution  of  con' 
struction  problems.  In  general,  to  investigate  the  solution 
of  a problem  by  this  method: 

Draw  a figure  in  which  the  required  construction  is 
assumed  as  made  ; 

Draw  auxiliary  lines , if  necessary  ; 

Observe  the  relations  between  the  -parts  of  this  figure,  in 
trder  to  discover  a Tcnown  relation  on  which  the  required 
construction  depends  ; 

Having  discovered  the  required  relation , construct  another 
figure  by  the  direct  rise  of  this  relation. 

Ex.  Through  a given  point  within  a circle  draw 
a chord  which  shall  be  bisected  by  the  given  point. 

Analysis.  Let  A be  the  given  point  within  the 
given  circle  0,  and  let  PQ  be  a chord  bisected  at 
the  point  A.  A bisected  chord  suggests  a line  OA 
joining  the  point  of  bisection  with  the  center  0, 
and  that  (Art.  224)  OA  ± PQ. 

Synthesis,  or  Direct  Solution.  Taking  another  figure  contain- 
ing the  data  of  the  problem,  connect  the  point  A with  the  center  of 
the  circle  by  the  line  OA. 

Through  A draw  a line  _L  OA  (Art.  274),  and  meeting  the  circumfer- 
ence at  the  points  P and  Q.  PQ  is  the  chord  required. 

EXERCISES.  CROUP  22 

CONSTRUCTION  OF  STRAIGHT  LINES. 

Ex.  1.  Draw  a line  parollel  to  a given  line,  and  tangent  to  a given 
eircle. 

[Sug.  Suppose  the  required  line  drawn;  then  the  radius  to  the 
point  of  tangency,  if  produced,  is  J.  given  line,  etc.] 

Ex.  2.  Draw  a line  perpendicular  to  a given  line,  and  tangent  to 
b given  circle. 


EXERCISES.  PROBLEMS 


167 


Ex.  3.  From  two  points  in  the  circumference  of  a circle,  draw 
two  equal  and  parallel  chords. 

Ex.  4.  Through  a given  point  draw  a line  which  shall  make  a 
given  angle  with  a given  line. 

Ex.  5.  Through  a given  point  draw  a line  which  shall  make  equal 
angles  with  the  sides  of  a given  angle. 

[Sug.  The  bisector  of  the  given  Z will  be  _L  the  required  line,  etc.] 

Ex.  6.  Through  a given  point  between  two  given  parallel  lines, 
draw  a line  of  given  length  with  its  extremities  in  the 
two  parallel  lines. 

Ex.  7.  Through  a given  point  A within  a circle, 
draw  a chord  equal  to  a given  line. 

Ex.  8.  From  a given  point  in  the  circumference 
of  a circle,  draw  a chord  at  a given  distance  from 
the  center. 

Ex.  9.  Through  a given  point  on  the  circumference  of  a circle, 
draw  a chord  which  shall  be  bisected  by  another  given  chord. 

[Sug.  Draw  the  radius  to  the  given  point  and  on  it  as  a diameter 
describe  a circle,  etc.  When  is  the  solution  impossible  f] 

294.  Construction  of  points  and  of  loci.  In  construct- 
ing a point  to  meet  certain  given  conditions,  it  is  often 
helpful  to  construct  the  locus  of  a point  answering  one  of  the 
given  conditions  and  observe  in  what  point  or  points  it  meets 
a given  line,  or  meets  another  locus  answering  another 
given  condition . 

EXERCISES.  CROUP  23 

CONSTRUCTION  OF  POINTS  AND  LOCI 

Ex.  1.  Find  a point  P in  a given  line  AB 
equidistant  from  two  given  points  C and  D.  V 

[Sug.  Construct  the  locus  of  all  points  & \ 

equidistant  from  C and  D and  observe  where  A -i— B 

it  intersects  the  given  line  AB.] 

Ex.  2.  Find  a point  P in  a given  circumference,  which  is  equi- 
distant from  two  given  points,  C and  D. 


168 


BOOK  n.  PLANE  GEOMETRY 


Ex.  3.  Find  a point  P in  a given  line,  which  is  equidistant  from 
two  given  intersecting  lines. 

Ex.  4.  Find  a point  in  a given  line,  which  is  at  a given  distance, 
d,  from  a given  point. 

Ex.  5.  Find  a point  which  is  at  a given  distance,  a,  from  a given 
point  A,  and  at  another  distance,  b,  from  another  given  point  B. 

Discuss  the  limitations  of  this  problem. 

Ex.  6.  Find  a point  equidistant  from  two  given  points,  and  at  a 
given  distance  from  a given  straight  line. 

[Sug.  Draw  the  locus  of  all  points  equidistant  from  the  two  given 
points,  and  also  the  locus  of  all  points  at  the  given  distance  from  the 
given  straight  line,  etc.] 

Ex.  7.  Find  a point  equidistant  from  two  given  points,  and  at  a 
given  distance  from  another  given  point. 

Ex.  8.  Find  a point  equidistant  from  two  given  points,  and  also 
from  two  given  intersecting  lines. 

On  the  other  hand,  the  determination  of  certain  loci  is 

equivalent  to  the  construction  of  all  points  which  satisfy  one 
or  more  given  conditions . 

Ex.  9.  Find  the  locus  of  the  center  of  a circle,  which  touches  a 
given  line  at  a given  point. 

[Sug.  Construct  a number  of  circles  touching  the  given  line  at 
the  given  point  and  observe  the  relation  of  their  centers.] 

Ex.  10.  Find  the  locus  of  the  center  of  a circumference  with  a 
given  radius,  r,  which  passes  through  a given  fixed  point. 

Ex.  11.  Find  the  locus  of  the  center  of  a circle,  touching  two  given 
intersecting  lines. 

Ex.  12.  Find  the  locus  of  the  center  of  a circle,  touching  two  given 
parallel  lines. 

Ex.  13.  Find  the  locus  of  the  center  of  a circle  of  given  radius,  r, 
which  touches  a given  straight  line. 

Ex.  14.  Find  the  locus  of  the  center  of  a circle  of  given  radius,  r, 
which  touches  a given  circle. 


EXERCISES.  PROBLEMS 


169 

EXERCISES.  GROUP  24 

CONSTRUCTION  OF  RECTILINEAR  FIGURES 

Construct 

Ex.  1.  An  equilateral  triangle,  given  the  perimeter. 

Ex.  2.  An  equilateral  triangle,  given  the  altitude. 

Ex.  3.  An  isosceles  triangle,  given  the  base  and  altitude. 

Ex.  4.  An  isosceles  triangle,  given  the  base  and  an  angle  at  the 
base. 

Ex.  5.  An  isosceles  triangle,  given  the  vertex  angle  and  the 
altitude. 

Ex.  6.  A right  triangle,  given  a leg  and  the  acute  angle  adjacent. 

Ex.  7.  A right  triangle,  given  a leg  and  the  acute  angle  opposite. 

Ex.  8.  A right  triangle,  given  the  hypotenuse  and  an  acute  angle. 

Ex.  9.  A right  triangle,  given  the  hypotenuse  and  a leg. 

Ex.  10.  A triangle,  given  the  altitude  and  the  sides  including  the 
vertical  angle. 

[Sug.  Through  the  foot  of  the  altitude  draw  a line  -L  altitude  and 
of  indefinite  length,  etc.] 

Ex.  11.  A triangle,  given  two  sides  and  the  altitude  upon  one  of 
them. 

Ex.  12.  A square,  given  the  diagonal. 

Ex.  13.  A rhombus,  given  the  two  diagonals. 

Ex.  14.  A rhombus,  given  one  angle  and  one  diagonal. 

Ex.  15.  A parallelogram,  given  two  adjacent  sides  and  an  altitude. 

Ex.  16.  A parallelogram,  given  a side,  the  altitude  upon  that  side 
and  an  angle. 

Ex.  17.  A parallelogram,  given  the  diagonals  and  an  angle 
included  by  them. 

Ex.  18.  A quadrilateral,  given  the  sides  and  one  angle. 


170 


BOOK  II.  PLANE  GEOMETRY 


295.  Use  of  auxiliary  lines  in  constructing  rectilinear 
figures.  In  constructing  polygons,  auxiliary  lines  are  fre- 
quently of  service.  Thus  it  is  often  of  especial  value  to 
construct,  Jirstl  either  the  inscribed  or  the  circumscribed 
circle,  and  afterward  the  required  triangle  or  quadrilateral . 

Ex.  Construct  an  isosceles  triangle,  given  the  base,  b,  and  the 
radius,  r,  of  the  inscribed  circle. 

Construction.  Draw  a circle  0 with  radius 
equal  to  r.  Draw  a tangent  at  any  point  A. 

On  this  tangent  mark  off  AB  and  AC  each 
equal  i b.  Prom  B and  C draw  tangents  BF 
and  CF  to  the  circle.  Then  BCF  is  the 
required  triangle. 


EXERCISES.  CROUP  25 

CONSTRUCTIONS.  AUXILIARY  LINES 

Construct 

Ex.  1.  An  isosceles  triangle,  given  the  base  and  the  radius  of  the 
circumscribed  circle. 

Ex.  2.  A right  triangle,  given  the  radius  of  the  circumscribed 
circle  and  one  leg. 

Ex.  3.  A right  triangle,  given  the  radius  of  the  circumscribed 
circle  and  an  acute  angle. 

Ex.  4.  A right  triangle,  given  the  radius  of  the  inscribed  circle 
and  an  acute  angle. 

[Sug.  Draw  the  inscribed  circle  and  at  its  center  construct  an 
angle  equal  to  the  supplement  of  the  given  angle.] 

Ex.  5.  A triangle,  given  the  base,  the  altitude  and  the  vertex 
angle. 

[Sug.  On  the  given  base  construct  a segment  which  shall  contain 
the  given  vertex  angle.  See  Art.  291.] 

Ex.  6.  A triangle,  given  the  base,  the  median  to  the  base,  and  the 
vertex  angle.  v 

Ex.  7.  A triangle,  given  one  side,  an  adjacent  angle,  and  the  radius 
of  the  inscribed  circle.  (See  Ex.  4.) 


EXEECISES.  PEOBLEMS 


171 


Ex.  8.  A triangle,  given,  one  side,  an  adjacent  angle,  and  the 
radius  of  the  circumscribed  circle. 

The  use  of  auxiliary  straight  lines  may  be  illustrated  as 
follows : 


Ex.  9.  Construct  a triangle,  given  the  perimeter  and  two  angles 

Analysis.  Suppose  the  required  triangle  ABC  already  constructed 
and  let  A ABC  and  ACB  be  the  given 
angles.  Produce  BC  to  D and  E , 
making  DB  = AB  and  CE  = AC. 

Then  BE=  given  perimeter.  Also 

IB  = ADAB  A ABC  = 2 ZB.  

Similarly  AACB=2AE.  Hence  D B C E 


Construction.  Take  BE  the  given  perimeter ; at  B construct  an 
angle  = /a  of  one  given  angle ; at  E construct  an  angle  = % of  the 
other  given  angle.  Produce  the  sides  of  these  angles  to  meet  at  A. 
Construct  A BAB  = AB  and  A CAE  = A E.  Then  A ABC  is  the 
required  triangle,  etc. 

Construct 

Ex.  10.  An  isosceles  triangle,  given  the  perimeter  and  the 
altitude. 

[Sug.  Bisect  the  perimeter  and  construct  the  altitude  A to  it  at 
its  midpoint.] 

Ex  11.  An  isosceles  triangle,  given  the  perimeter  and  the  vertex 
angle. 

[Sug.  If  the  vertex  Z is  known,  the  base  A may  be  obtained.] 

Ex.  12.  A right  triangle,  given  an  acute 
angle  and  the  sum  of  the  legs. 

[Sug.  Given  AB  the  sum  of  the  legs, 
construct  ZA=45°,  etc.] 

Ex.  13.  A right  triangle,  given  an  acute  angle  and  the  difference 
of  the  legs. 

Ex.  14.  A right  triangle,  given  the  hypotenuse  and  the  sum  of 
the  legs. 

Ex.  15.  A right  triangle,  given  an  acute  angle  and  the  sum  of  the 
hypotenuse  and  gne  leg, 


172 


BOOK  II . PLANE  GEOMETRY 


Ex.  16.  A triangle,  given  an  angle,  a side  and  the  sum  of  the 
other  two  sides. 

Ex.  17.  A triangle,  given  an  angle  A,  the  sum  of  the  sides  AB 
and  BC  and  the  altitude  upon  AB. 

296.  Reduction  of  problems.  In  many  cases  a problem 
may  be  solved  by  reducing  the  problem  to  a problem  already 
solved.  (This  is  a special  kind  of  analysis.) 

Ex.  Construct  a parallelogram,  given  the  diagonals  and  one  side. 

Analysis:  Suppose  the  07  ABCF  to  he  the  required  07  already 
constructed.  Let  AF  he  the  given  side.  If  the  diagonals  are 
given,  half  each  diagonal  is  given  (Art. 

161).  Hence  in  the  A AOF  the  three 
sides  are  given.  Hence  the  required 
problem  reduces  to  the  problem  of  con- 
structing a triangle  whose  three  sides 
are  given  (Art.  283).  Hence 

Construction.  Let  the  pupil  supply  the  direct  construction. 


EXERCISES.  CROUP  26 

REDUCTION  OF  CONSTRUCTION  PROBLEMS 

Construct 

Ex.  1.  A right  triangle,  given  the  altitude  upon  the  hypotenuse 
and  the  median  upon  the  same. 

Ex.  2.  A rectangle,  given  the  perimeter  and  a diagonal  (see 
Ex.  14,  p.  171). 

Ex.  3.  A rectangle,  given  the  perimeter  and  an  angle  made  by 
the  diagonals. 

Ex.  4.  A triangle,  given  the  three  angles  and  the  radius  of  this, 
circumscribed  circle. 

[Sug.  The  sides  of  the  A are  the  chords  of  the 
segments  of  the  O containing  the  given  A .] 

Ex,  5.  A triangle,  given  two  sides  and  the  median  . , 

j / 

to  the  third  side.  K 


EXERCISES.  PROBLEMS 


173 


Ex.  6.  A triangle,  given  the  three  medians. 

[Sug.  Reduce  this  to  the  preceding  Ex.  See  Art.  187.] 

Ex.  7.  An  isosceles  trapezoid,  given  the  bases  and  an  angle. 

Ex.  8.  An  isosceles  trapezoid,  given  the  bases  and  a diagonal. 

Ex.  9.  A trapezoid,  given  the  four  sides. 

Ex.  10.  A trapezoid,  given  the  bases  and  the  two  diagonals. 

[Sug.  Reduce  to  Art.  283  by  producing  the  lower  base  a distance 
equal  to  the  upper  base,  etc.] 

297.  Construction  of  circles.  The  construction  of  a 
required  circle  is  frequently  a good  illustration  of  the 
preceding  method  of  reducing  one  construction  problem  to 
another.  For  the  construction  of  a circle  frequently  re- 
duces to  the  problem  of  finding  a point  ( the  center  of  the 
circle ) which  answers  given  conditions.  (See  Art.  294.) 

Ex.  Construct  a circle  which  shall  touch  two  given  intersecting 
lines  and  have  its  center  in  another  given  line. 

This  problem  is  equivalent  to  the  problem  of  finding  a point  which 
shall  be  in  a given  line  and  be  equidistant  from  two  other  given 
lines.  (See  Ex.  3,  p.  168.)  In  some  cases,  however,  the  construc- 
tion of  a required  circle  must  be  made  by  an  independent  method. 

EXERCISES.  CROUP  27 

CONSTRUCTION  OP  CIRCLES 
Construct  a circle  with  given  radius,  r, 

Ex.  1 . Which  passes  through  a given  point  and  touches  a given 
line. 

Ex.  2.  Which  has  its  center  in  a given  line  and  touches  another 
given  line. 

Ex.  3.  Which  passes  through  two  given  points. 

Construct  a circle 

Ex.  4.  Which  touches  two  given  parallel  linos  and  passes  through 
a given  point. 


174 


BOOK  H.  PLANE  GEOMETRY 


Ex.  5.  Which  passes  through  two  given  points  and  has  its  center 
on  a given  line. 


Ex.  6.  Which  touches  three  given  lines, 
two  of  which  are  parallel. 

Ex.  7.  Which  passes  through  a given 
point  A and  touches  a given  line  BC  at  a 
given  point  B. 


[Sug.  Draw  AB  and  at  B construct  a J. 
to  PC.] 

Ex.  8.  Which  touches  a given  line  and  also 
touches  a given  circle  at 
a given  point  A. 

Ex.  9.  Which  touches  a given  line  AB 
_j5  at  a given  point  A and  touches  a given 
circle. 


EXERCISES.  CROUP  28 

PROBLEMS  SOLVED  BY  VARIOUS  METHODS 

Ex.  1.  Through  a given  point  draw  a line  which  shall  cut  two 
given  intersecting  lines  so  as  to  form  an  isosceles  triangle. 

Ex.  2.  Construct  an  isosceles  triangle,  given  the  altitude  and 
one  leg. 

Ex.  3.  In  a given  circumference  find  a point  equidistant  from  two 
given  intersecting  lines. 

Ex.  4.  Draw  a circle  which  shall  touch  two  given  intersecting 
lines,  one  of  them  at  a given  point. 

Ex . 5.  Draw  a line  which  shall  be  terminated  by  the  sides  of  a given 
angle,  shall  equal  a given  line,  and  be  parallel  to  another  given  line. 

Ex.  6.  Construct  a triangle,  given  one  side,  an  adjacent  angle, 
and  the  difference  of  the  other  two  sides. 

Ex.  7.  Find  a point  in  a given  circumference  at  a given  distance 
from  a given  point. 


MISCELLANEOUS  EXEECISES.  PEOBLEMS  175 


Ex.  8.  Construct  a parallelogram,  given  a side,  an  angle,  and  a 
diagonal. 

Ex.  9.  Through  a given  point  within  an  angle^  draw  a straight  line 
terminated  by  the  sides  of  the  angle  and  bisected  by  the  given  point. 

[Sug.  Draw  a line  from  the  vertex  of  the  angle  to  the  given  point 
and  produce  it  its  own  length  through  the  point.] 

Ex.  10.  Construct  a triangle,  given  the  vertex  angle  and  the 
segments  of  the  base  made  by  the  altitude. 

[Sug.  Use  Art.  291.] 

Ex.  11.  Construct  an  isosceles  triangle,  given  the  angle  at  the 
vertex  and  the  base. 

Ex.  12.  Draw  a circle  with  given  radius  which  shall  touch  a given 
circle  at  a given  point. 

Ex.  13.  Construct  a right  triangle,  given  the  hypotenuse  and  the 
altitude  upon  the  hypotenuse. 

Ex.  14.  Construct  a triangle,  given  the  base  and  the  altitudes 
upon  the  other  two  sides. 

[Sug.  Construct  a semicircle  on  the  given  base  as  a diameter.] 

Ex.  15.  Find  a point  in  one  side  of  a triangle  equidistant  from 
the  other  two  sides. 

Ex.  16.  Construct  a triangle,  given  the  altitude  and  the  angles  at 
the  extremities  of  the  base. 

Ex.  17.  Construct  a rhombus,  given  an  angle  and  a diagonal. 

Ex.  18.'  Draw  a circle  which  shall  pass  through  two  given  points 
and  have  its  center  equidistant  from  two  given  parallel  lines. 

Ex.  19.  Construct  a triangle,  given  one  side,  an  adjacent  angle 
and  the  radius  of  the  circumscribed  circle. 

Ex.  20.  In  a given  circle  draw  a chord  equal  to  a given  line  and 
parallel  to  another  given  line. 

[Sug.  Find  the  distance  of  the  given  chord  from  the  center,  by 
constructing  a right  triangle  of  which  the  hypotenuse  and  one  leg  are 
given.] 


176 


BOOK  II.  PLANE  GEOMETRY 


Ex.  21.  Construct  a triangle,  given  an  angle,  the  bisector  of  that 
angle,  and  the  altitude  from  another  vertex. 

Ex.  22.  Find  the  locus  of  the  points  of  contact  of  tangents  drawn 
from  a given  point  to  a series  of  circles  having  a given  center. 

[Sug.  Use  Arts.  229  and  261.] 

Ex.  23.  Given  aline  AB  and  two  points. 

C and  D,  on  the  same  side  of  AB.  Find  a 
point  P in  AB  such  that  Z.APC=  ZBPD. 

[Sug.  Draw  a X from  C to  AB  and  pro- 
duce it  its  own  length,  etc.] 

Ex.  24.  Given  a line  AB  and  two  points  C and  D on  the  same  side 
of  AB;  find  a point  P in  AB  such  that  CP  + PJD  shall  be  a minimum. 

Ex.  25.  Draw  a common  external 
tangent  to  two  given  circles. 

Ex.  26.  Draw  a common  internal 
tangent  to  two  given  circles. 


Book  III 


PROPORTION.  SIMILAR  POLYGONS 

THEORY  OF  PROPORTION 

298.  Ratio  has  been  defined,  and  its  use  briefly  indi- 
cated in  Arts.  245,  246. 

299.  A proportion  is  an  expression  of  the  equality  of 

two  or  more  equal  ratios.  As, 

a c _ _ 

- =-,  or  a : o = c : a. 

I d 

This  reads,  "the  ratio  of  a to  b equals  the  ratio  of  c to 
d”  or,  "a  is  to  6 as  c is  to  d.” 

300.  The  terms  of  a proportion  are  the  four  quantities 
used  in  the  proportion.  In  a proportion 

the  antecedents  are  the  first  and  third  terms; 

the  consequents  are  the  second  and  fourth  terms; 

the  extremes  are  the^rsi  and  last  terms; 

the  means  are  the  second  and  third  terms. 

A fourth  proportional  is  the  last  term  of  a proportion 
(provided  the  means  are  not  equal). 

Thus,  in  a : b = c : d,  d is  a fourth  proportional. 

301.  A continued  proportion  is  a proportion  in  which 
each  consequent  and  the  next  antecedent  are  the  same. 

Thus,  a : b = h :c  = c : d = d : e is  a continued  proportion. 

A mean  proportional  is  the  middle  term  in  a continued 
proportion  containing  but  two  ratios. 

(177) 


L 


178 


BOOK  III.  PLANE  GEOMETRY 


A third  proportional  is  the  last  term  in  a continued  pro- 
portion containing  but  two  ratios. 

Thus,  in  a : b = b : c,  b is  a mean  proportional,  andc  is  a 
third  proportional. 


Proposition  I.  Theorem 

302.  In  any  proportion , the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Given  the  proportion  a : b = c : d. 

To  prove  ad  — he. 

Proof.  p = C~.  • Hyp. 

h d 

Multiply  each  member  by  hd. 

ad  = be.  As.  4. 

Q.  E.  D. 


Proposition  II.  Theorem 


303.  The  mean  proportional  between  two  quantities  is 
equal  to  the  square  root  of  their  product. 

Given  the  proportion  a : b = b : c. 

To  prove  b = V^ac. 

Proof.  a : b = b : c.  Hyp. 

.*.  b2  = ac,  Art.  302. 

{in  any  proportion,  the  product  of  the  extremes  equals  the  product  of  the 

means). 


Z>  = ‘ 


ac. 


Q.  E.  D. 


Ex.  1.  Find  the  fourth  proportional  to  2,  3 and  6;  also  to  3,  i,  f. 
Ex.  2.  Find  the  mean  proportional  between  3 and  6. 

Ex.  3,  Find  the  third  proportional  to  3 and  5. 


THEORY  OF  PROPORTION 


179 


Proposition  III.  Theorem 

304.  If  the  product  of  two  quantities  is  equal  to  the  prod- 
rid  of  two  other  quantities,  one  pair  may  be  made  the  ex- 
tremes and  the  other  pair  the  means  of  a proportion. 

Given  ad  = bc. 


To  prove 
Proof. 

Divide  each  member 

Then 

Or 


a : b=*c  : d. 


ad— be. 

by  bd. 

a _ c 
b d 

a : b — c : d. 


Hyp. 


Ax.  5. 


Q.  E.  D. 


305.  Cor.  1.  If  the  antecedents  of  a proportion  are 
equal,  the  consequents  are  equal. 

Thus,  if  a-.x  — a-.y,  then  x—y. 

Let  the  pupil  supply  the  proof. 


306.  Cor.  2.  If  three  terms  of  one  proportion  are 
equal  to  the  corresponding  three  terms  of  another  proportion , 
the  fourth  terms  of  the  two  proportions  are  equal. 

Thus,  if  a : b = c : x,  and  a : b = c : y,  then  x = y. 

Let  the  pupil  supply  the  proof. 


Ex.  1.  Write  ab=pq  as  a proportion  in  as  many  different  ways  as 
possible. 

Ex.  2,  Write  *(»+l)~6  as  a proportion;  also 


180 


BOOK' III.  PLANE  GEOMETRY 


Proposition  IV.  Theorem 

307.  If  four  quantities  are  in  proportion , they  are  in 
proportion  by  alternation ; that  is,  the  first  term  is  to  the 
third  as  the  second  is  to  the  fourth. 

Given  the  proportion  a : b = c : d. 

To  prove  a : c = b : d. 

Proof.  a : b = c : d.  Hyp. 

.'.  ad=bc,  Art.  302. 

(in  any  proportion,  the  product  of  the  extremes  equals  the  product  of 

the  means) . 

Writing  a and  d as  the  extremes,  and  c and  b as  the 
means  of  a proportion, 

a : C=b  : d,  Art.  304. 

(if  the  product  of  two  quantities  is  equal  to  the  product  of  two  other 

quantities,  one  pair  may  he  made  the  extremes  and  the  other  pair 
the  means  of  a proportion) . 

Q.  E.  D. 


Proposition  V.  Theorem 

308.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  inversion ; that  is,  the  second  term  is  to  the 
first  as  the  fourth  is  to  the  third. 

Given  the  proportion  a :b=c  : d. 

To  prove  b : a — d : c. 

proof.  a : b = c : d.  Hyp 

.*.  ad  = bc.  (Why?) 

Writing  b and  c as  the  extremes,  and  a and  d as  the 
means, 

b : a — d : C.  (Why?) 

Q.  E.  B. 


Ex.  Transform  x:a=h;c  so  that  x shall  be  the  last  term. 


THEORY  OF  PROPORTION 


181 


Proposition  VI.  Theorem 


309.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  composition;  that  is,  the  sum  of  the  first  two 
xerms  is  to  the  second  term  as  the  sum  of  the  last  two  terms 
is  to  the  last  term. 


Given  the  proportion  a : b = c : d. 

To  prove  a + b : b — cff  d : d. 


Proof. 


Add  1 to  each  member  of  the  equality. 

1+H+1’ 


Hyp. 
Ax.  2. 


Or 


a~\-  b c d 

~b  ~7l 


That  is  a + b : b = c + d : d. 

Let  the  pupil  show  also  that  a-fb  : « = c + d : c. 

Q.  E.  D. 


Proposition  VII.  Theorem 

310.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  division ; that  is,  the  difference  of  the  first 
two  is  to  the  second  as  the  difference  of  the  last  two  is  to 
the  last. 

Given  the  proportion  a : b = c : d. 

To  prove  a — b : b = c — d : d. 

a _c_ 

b~d' 


Proof. 


Hyp. 


182 


BOOK  III.  PLANE  GEOMETRY 


Subtract  1 from  each  member  of  the  equality. 

1. 

a 

a — b c — d 


a _ c 

A b d 


Ax.  3. 


Or 

That  is 


b d 

a — b : b = c — d : d.  q.  e.  i». 

Let  the  pupil  show  also  that  a — b : a = c — d : c. 


Proposition  VIII.  Theorem 

311.  If  four  quantities  are  in  proportion , they  are  in 
proportion  by  composition  and  division;  that  is,  the  sum  of 
the  first  two  is  to  their  difference  as  the  sum  of  the  last  two 
is  to  their  difference. 

Given  the  proportion  a : b = c •.  d. 

To  prove  a fi-  b : a — b = c fi-  d : c — d. 


Proof. 

II 

e 

Hyp. 

By  composition 

a -\-  b cfi-  d 

Art.  309. 

b d 

Also  by  division 

a — b _ c — d 

Art.  310 

b d 

Dividing  the  corresponding  members  of 

the  two 

equalities, 

a b c fir  d 

Ax.  5. 

a — b c — d 

That  is  a + b : 

: a — b — cfi-d ■ c — d. 

Q.  E.  D. 

Ex.  1.  What  does  the  proportion  12:3  = 8:2  become  by  composi- 
tion? also  by  division? 

Ex.  2,  What  does  2x — 5:5=3x — 7:7  become  by  composition? 


THEOET  OF  PEOPOKTION 


183 


Pkoposition  IX.  Theorem 

312.  In  a series  of  equal  ratios , the  sum  of  all  the  ante - 
cedents  is  to  the  sum  of  all  the  consequents  as  any  one  ante- 
cedent is  to  its  consequent. 

Given  a : b — c : d = e : /=  g : h. 

To  prove  a + c + e + g : b + d +/+  h = a : b. 

Proof.  Denote  each  of  the  equal  ratios  by  r. 

Then  7 = r,  a = br. 

b 

Similarly  c — dr,  e=fr,  g = hr.  Ax.  4. 

Hence  a-\-c-\-e-\-g=(b-\-d-\-f-\-li)  r.  Ax.  2. 

Dividing  by  b 4 d +/+  h,  — y ^ = r = - ■ Ax.  5. 

That  is  a + c + e + g : b + d + / + h = a : b . 

Q.  E.  D. 


Proposition  X.  Theorem 

313.  The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 

Given  a : b — c : d,  e : f=g  : h,  and  j : Tc=l  : m. 

To  prove  aej  : bfk  — cgl  : dhm. 

r,  , a _ c e _ g j _ l 

Proof.  7=7,  7— — • Hyp. 

b d j h k m 

Multiplying  together  the  corresponding  terms  of  these 
equalities, 

aej  _ cgl 
bfk  dhm 

aej  : bfk  — cgl  : dhm. 


Ax.  4. 


That  is 


Q.  E.  D. 


184 


BOOK  III.  PLANE  GEOMETPtY 


Proposition  XI.  Theorem 


314.  Like  poivers  or  like  roots  of  the  terms  of  a propor • 
Hon  are  in  proportion. 


Given  the  proportion  a : b — c : d. 

To  prove  an  : bn  = cn  : dn,  and  a»  ; b"  = c"  : 

Proof.  7=  -•  Hyp 

o d 

Raising  both  members  to  the  nth  power, 
an  cn 


bn  dn 

That  is  an  : bn  = cn  : dn  . 

In  like  manner  a » : bTt  = c”  : d 


Q.  E. 


Proposition  XII.  Theorem 


315.  Equimultiples  of  two  quantities  have  the  same 
ratio  as  the  quantities  themselves. 


Given  the  two  quantities  a and  b. 

To  prove  ma  : mi=a  : b. 

_ . a a 

Proof.  -•  Ident. 

0 b 


Multiply  each  term  of  the  first  fraction  by  m. 
.ma  _ a 
mb  b 

That  is  ma  : mb  — a : b. 


Q.  E.  D. 


Ex.  1.  Transform  p : q=x  : y in  all  possible  ways  by  the  use  of  the 
properties  of  a proportion. 

Ex.  2.  Transform  7 :3=28:12  by  composition  and  division. 

Ex.  3.  Also  2a;  + 5: 2x— 6 = x3+l;  ff2  — 1- 


THEORY  OF  PROPORTION 


185 


316.  Note.  In  the  theory  of  proportion  as  just  presented,  the 
quantities  used  are  assumed  to  be  commensurable,  but  the  same 
theorems  may  also  be  proved  for  proportions  whose  terms  are  incom- 
mensurable by  use  of  the  method  of  limits.  For  each  incommensu- 
rable ratio  may  be  made  the  limit  of  a corresponding  commensurable 
ratio ; then,  by  showing  that  the  variable  commensurable  ratios  are 
equal,  it  may  be  proved  that  the  limiting  incommensurable  ratios 
are  equal. 

It  is  also  to  be  noted,  that,  in  the  above  theorems,  the  terms  of  a 
ratio  must  be  of  the  same  kind  of  quantity;  that  is,  both  be  lines,  or 
both  be  surfaces,  etc.  Hence,  in  order  that  a proportion  be  treated  by 
alternation,  for  instance,  all  four  of  the  terms  must  be  of  the  same 
kind. 


Ex.  1.  Find  the  fourth  proportional  to  a,  2 a,  3r. 

Ex.  2.  Find  the  third  proportional  to  a + b and  a — 6. 

Ex.  3.  Find  the  value  of  x in  the  proportion,  4 : 5=x  : 15. 

Ex.  4.  Find  a short  method  of  determining  whether  a given  propor- 
tion is  true  or  not.  Use  this  method  to  determine  whether  the  follow- 
ing proportions  are  true : (1)4:  6 = 3:  9.  (2)  5 a : 2 a=15  : 6. 

Ex.  5.  Transform  the  following  proportion  by  composition  and 
division,  and  afterward  find  the  value  of  a;;  *+  1 : x — 1 = 7 : 5. 

Ex.  6.  Construct  exactly  the  figure  of  page  77  with  the  fewest  pos- 
sible adjustments  of  the  compasses. 

Ex.  7.  Draw  an  obtuse  triangle  and  construct  its  three  altitudes, 
using  the  concurrence  of  the  altitudes  as  a test  of  the  accuracy  of  the 
work. 

Ex.  8.  Arrange  nine  points  in  a plane  in  such  a way  that  the 
greatest  number  of  straight  lines  may  pass  through  them,  each  line  tc 
pass  through  three  points  and  only  three. 


186 


BOOK  in.  PLANE  GEOMETRY 


Proposition  XIII.  Theorem 

317.  A line  parallel  to  one  side  of  a triangle  and  meet 
ing  the  other  two  sides , divides  these  sides  proportionally . 

A A 


Given  the  triangle  ABC  and  the  line  I)E  ||  base  BG  and 
intersecting  the  sides  AB  and  AC  in  the  points  I)  and  E, 
respectively. 

To  prove  T)B  : AD  = E^ : AE. 

Case  I.  When  DB  and  AD  (Fig.  1)  are  commensurable . 

Proof.  Take  any  common  unit  of  measure  of  DB  and 
AD,  as  AK,  and  let  it  be  contained  in  DB  a certain  number 
of  times,  as  n times,  and  in  AD,  rn  times. 

Then  ^ = -•  Art-  24A 

AD  m 

Through  the  points  of  division  of  DB  and  AD  draw 
lines  ||  BG. 

These  lines  will  divide  EG  into  n,  and  AE  into  m parts, 
all  equal.  Art.  176. 

. E - V (Why?) 

AE  m 

DB  EG 

•’*  AD  AE 


(Why  ?) 


PROPORTIONAL  LINES 


187 


Case  II.  When  DB  and  AD  (Fig.  2)  are  incommensurable. 
Let  the  line  AD  be  divided  into  any  number  of  equal 
parts  and  let  one  of  those  parts  be  applied  to  DB.  It  will 
be  contained  in  DB  a certain  number  of  times,  with  a line 
PB,  less  than  the  unit  of  measure,  as  a remainder. 

Draw  PQ\\  BC. 

Then  DP  and  AD  are  commensurable.  Gonstr. 


DP  _EQ 
•’*  AD  AE 


Case  I. 


If  now  the  unit  of  measure  be  indefinitely  diminished, 
the  line  PB , which  is  less  than  the  unit  of  measure,  will  be 
indefinitely  diminished. 

Hence  DP  = DB,  and  EQ  A EC  as  a limit.  Art.  251. 


Hence  becomes  a variable  with  yy  as  its  limit. 

AD  AD  Art.  253,  3, 

E 0 EG 

Also  -rA  becomes  a variable  with  -p=  as  its  limit. 

AE  AE  Art.  253,  3. 

DP  E 0 

But  the  variable  — — = the  variable  ~r~  always.  Case  I. 
AD  AE 


the  limit  -y^  = the  limit 

AD  AE 


Art.  254. 

Q.  E.  D. 


318.  Cor.  1.  By, composition,  Art.  309. 

DB  + AD  : AD  = EC  A AE  -.  AE. 

Or  AB  : AD  = AG  : AE. 

In  like  manner  AB  : DB  = AG  : EC, 
oy,  in  general  language,  if  a line  parallel  to  the  base  cut  the 
sides  of  a triangle,  a side  is  to  a segment  of  that  side  as  the 
other  side  is  to  the  corresponding  segment  of  the  second  side. ' 


319.  Cor.  2.  Using  Fig.  2, 

PB=AP  DP=AP  PB  DP 
QG  AQ ; alS°  EQ  AQ  " QC  EQ 
Hence,  if  two  lines  are  cut  by  a number  of  parallels,  the 
corresponding  segments  are  proportional. 


188 


BOOK  in.  PLANE  GEOMETRY 


Proposition  XIV.  Theorem  (Converse  of  Prop.  XIII) 

320.  If  a straight  line  divides  two  sides  of  a triangle 
proportionally , it  is  parallel  to  the  third  side. 


A 


Given  the  A ABC  and  the  line  DF  intersecting  AB  and 
AC  so  that  AB  : AD  — AC  : AF. 

To  prove  DF  ||  BC. 

Proof.  Through  D draw  the  line  DK  ||  BC  and  meeting 
the  side  AC  in  K. 

Then  AB  : AD  = AC:  AK,  Art.  313. 

(if  a line  ||  base  cut  the  sides  of  a A,  a side  is  to  a segment  of  that  side  as 
the  other  side  is  to  the  corresponding  segment  of  the  second  side). 

But  AB  : AD  — AC  : AF.  Hyp. 

.-.  AF=AK,  Art.  306. 

(if  three  terms  of  one  proportion  are  equal  to  the  corresponding  three 
terms  of  another  proportion,  the  fourth  terms  of  the  two 
proportions  are  equal). 

Hence  the  point  K falls  on  F,  and  the  line  DK  coin- 
cides with  the  line  DF.  Art.  66. 

But  the  line  DK  ||  BC.  Constr 

.-.  DF  ||  BC, 

( for  DF  coincides  with  DK,  which  is  ||  BC). 

Q.  E.  D. 

Ex.  1„  In  the  figure  of  Prop.  XIII,  if  AD=6,  DB= 4,  and  AE=9, 
find  EC. 

Ex.  2.  Also,  if  AD= 12,  D£  = 8,  and  AC=  15,  find  AE  en"  EC. 


EXERCISES 


189 


EXERCISES.  CROUP  29 

REVIEW  EXERCISES 
Make  a list  of  the  properties  of 

Ex.  1.  One  straight  line  (in  connection  with  such  points  as  may 
he  related  to  the  line). 

Ex.  2.  Two  straight  lines  that  meet,  or  intersect  (in  connection 
with  such  angles  as  may  be  formed  by  the  given  lines). 

Ex.  3.  Two  or  more  parallel  lines  (in  connection  with  transversals 
and  angles  formed). 

Ex.  4.  Right  angles. 

Ex.  5.  Complementary  angles. 

Ex.  6.  Supplementary  angles. 


Ex,  7.  A single  triangle  (in  connection  with  lines  within  the 
triangle,  as  altitudes,  medians,  etc.). 


Ex.  8.  A right  triangle. 

Ex.  9.  An  isosceles  triangle. 
Ex.  10.  Two  triangles. 

Ex.  11.  Two  right  triangles. 
Ex.  12.  A quadrilateral. 


Ex.  13.  A trapezoid. 

Ex.  14.  An  isosceles  trapezoid. 
Ex.  15.  A parallelogram. 

Ex.  16.  A rhombus. 

Ex.  17.  A polygon. 


Ex.  18.  A single  circle,  or  circumference  (in  connection  with 
related  points). 

Ex.  19.  Arcs  of  a circle  (in  connection  with  related  lines  and 
rngles) . 

Ex.  20.  Chords  in  a circle  (in  connection  with  related  lines 
or  arcs). 

Ex.  21.  Central  angles  in  a circle. 

Ex.  22.  Tangents  to  a circle. 

Ex.  23.  Secants  to  a circle 

Ex.  24.  Two  circles. 


190 


BOOK  III.  PLANE  GEOMETRY 


SIMILAR  POLYGONS 


321.  Def.  Similar  polygons  are  polygons  having  their 
homologous  angles  equal  and  their  homologous  sides  pro- 
portional.  b 


E D E' 

Thus,  if  the  figures  ABODE  and  A'B'G'D'E'  are  similar 
the  angles  A,  B,  G,  etc.,  must  equal  the  angles  A' , B' , C' , etc., 
respectively;  also  AB  A'B'  — BC  : B'G'—CD  : C'D' , etc. 

Hence  it  is  constantly  to  he  borne  in  mind  that  simi- 
larity in  shape  or  form  of  rectilinear  figures  involves  two 
distinct  pi’operties: 

1.  The  homologous  angles  are  equal. 

2.  The  homologous  sides  are  proportional. 

It  should  also  be  clearly  realized  that  one  of  these 
properties  may  be  true  of  two  figures,  ana  not  the  other. 

Thus,  in  the  rectangle  A and  the  rhomboid  B,  the  cor- 
responding sides  are  proportional  but  the  corresponding 
angles  are  not  equal. 


D 


Also,  in  the  rectangle  G and  the  square  D,  the  corres- 
ponding angles  are  equal  but  the  corresponding  sides  are 
not  proportional. 

However,  it  will  be  found  that,  in  the  case  of  triangles,  if 
one  of  the  two  properties  is  true,  the  other  must  be  true  also. 


322.  Def.  The  ratio  of  similitude  in  two  similar  figures 
is  the  ratio  of  any  two  homologous  sides  in  those  figures. 


SIMILAR  POLYGONS 


191 


Proposition  XV.  Theorem 

323.  If  two  triangles  are  mutually  equiangular , they 
are  similar. 

A 


Given  the  A ABC  and  A'B'C'  with  Z A = Zl',  ZB  = 
ZB',  and  ZC  = ZC . 

To  prove  the  A ABC  and  A'B'C'  similar. 

Proof.  The  given  A are  mutually  equiangular.  Hyp. 

Hence  it  only  remains  to  prove  that  their  homologous 
sides  are  proportional.  Art.  S21. 

Place  the  A A'B'C'  upon  the  A ABC , so  that  /.A'  shall 
coincide  with  its  equal,  the  Zl,  and  B'C  take  the  posi- 
tion FE. 


Then 

ZAFE  = ZB. 

Hyp. 

:.  FEW  BC. 

(Why  ?) 

:.  AB  : AF=AC  : AE. 

Art.  317. 

Or 

AB  : A'B'=AC  : A’C. 

Ax.  8. 

In  like 

manner,  by  placing  the  A A'B'C' 

upon  the  A 

ABC  so  that  the  ZB'  shall  coincide  with  its  equal,  the  IB, 
it  may  be  proved  that  AB  : A'B'  = BC  : B'C'. 

Hence  the  A ABC  and  A'B'C'  are  similar.  Art.  321. 

324.  Cor.  1.  If  two  triangles  have  two  angles  of  one  equal 
to  two  angles  of  the  other,  the  triangles  are  similar;  also, 

If  two  right  triangles  have  an  acute  angle  of  one  equal  to 
an  acute  angle  of  the  other,  the  triangles  are  similar. 

325.  Cor.  2.  If  two  triangles  are  each  similar  to  the 
same  triangle,  they  are  similar  to  each  other. 


192 


BOOK  ni.  PLANE  GEOMETRY 


Proposition  XVI.  Theorem 


326.  If  two  triangles  have  their  homologous  sides  pro- 
portional, they  are  similar. 


Given  tlie  A ABC  and  A'B'C , in  which  AB  : A'B'  = 
AC  : A'C'  = BC  : B'C'. 

To  prove  the  A ABC  and  A'B'C'  similar. 

Proof.  The  given  A have  their  homologous  sides  pro- 


portional. Hyp. 

Hence  it  only  remains  to  prove  that  the  A are  mutually 
equiangular.  Art.  321. 

1.  On  AB  take  AF  equal  to  A'B',  and  on  AC  take  AH 
equal  to  A'C'.  Draw  FH. 

Then  AB  : AF=AC  : AH.  Hyp. 

.*.  FH  ||  BC,  Art.  320. 

'if  a straight  line  divides  two  sides  of  a A proportionally , it  is  ||  the 

thud  side) . 

A Z AFH  = IB,  and  Z AHF  = Z C.  (Why?) 

A AFH  and  ABC  are  similar,  Art.  323. 

(if  two  A are  mutually  equiangular,  they  c re  similar) . 

2.  .'.  AB  : AF=BC  : FH.  Art.  321. 

Or  AB  : A'B'  = BC  : FH.  As.  8. 

But  AB  : A'B' — BC  ■ B'C',  Hyp. 


SIMILAR  POLYGONS 


193 


:.  FH=B'C'.  Art.  306. 

Hence  the  A AFH  and  A'B'C'  are  equal.  Art.  101. 

A A'B'C'  is  similar  to  A ABC , Ax.  8 

( for  its  equal  A AFH  is  similar  to  A ABC). 

Q.  E.  I>. 

Proposition  XVII.  Theorem 

327.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other , and  the  including  sides  proportional,  the 
triangles  are  similar. 

A 

A‘ 


B G B'  O' 

Given  the  A ABC  and  A'B'C',  in  which  Zl=  A A'  and 
AB  : A'B'  = AC  : A'C. 

To  prove  the  A ABC  and  A'B'C'  similar. 

Proof.  Place  the  A A'B'C'  upon  the  A ABC  so  that 
A A'  shall  coincide  with  its  equal,  the  A A,  and  B'C  take 


the  position  FH. 

Then  AB  -.  AF=AC : AH.  Hyp. 

Hence  FH  ||  BC.  (Why?) 

/.  AAFH=  AB,  and  AAHF—AC.  (Why?) 
.'.  A ABC  and  AFH  are  similar.  Art.  323. 
Or  A ABC  and  A'B'C'  are  similar.  Ax.  8. 


Q.  E.  D. 

Ex.  In  the  A ABC,  AB= 6,  AC= 8,  BC<=  10;  in  the  similar  A A'BfCt 
A'B'=9.  Find  A'C ' and  B'C. 

M 


194 


BOOK  III.  PLANE  GEOMETRY 


Proposition  XVIII.  Theorem 

328.  If  two  triangles  have  their  sides  parallel , or  per * 
pendicular , each  to  each,  the  triangles  are  similar. 


C' 
Fig.  3 


Given  the  A A'B'O  (Fig.  2)  with  its  sides  ||,  and  the 
A A'B'O  (Fig.  3)  with  its  sides  J_,  to  corresponding  sides 
of  the  A ABC. 

To  prove  A ABC  and  A'B'C'  similar. 

Proof.  The  A A and  A'  are  either  equal  or  supple- 
mentary, ( for  the  sides  forming  them  are  ||  or  X).  Arts.  130, 132. 

Similarly,  the  A B and  B' , and  C and  O are  either 
equal  or  supplemental’}*. 

Hence  one  of  the  three  following  statements  must  be 
true  concerning  the  angles  of  the  A : either 

1.  The  A contain  three  pairs  of  supplementary  A and 
A A A A A' = 2 rt.  A,  AB  + AB'  = 2 rt.  A,  AC  A AC' = 
2 rt.  A ; or 

2.  The  A contain  two  pairs  of  supplementary  A , as 
AA=AA',  AB-\-AB'  = 2 rt.  i,  AC  A AO  = 2 rt.  A;  or 

3.  The  A contain  three  pairs  of  equal  A and  AA  = 

AA',  AB=  AB',  AC=  AC',  Art.  139. 

(if  two  A of  a A = two  A of  another  A,  the  third  A are  equal). 

The  first  two  of  these  statements  are  impossible,  for  the 
sum  of  the  A of  two  A cannot  exceed  four  rt.  A . Art.  134. 

Hence  the  third  statement  is  true,  and  the  A ABC  and 
A'B'O  are  mutually  equiangular,  and  therefore  similar. 

Art.  323. 

Q.  E.  D. 


SIMILAR  POLYGONS 


195 


Proposition  XIX.  Theorem 

329.  If  two  polygons  are  similar , they  may  he  separated 
into  the  same  number  of  triangles,  similar,  each  to  each,  and 
similarly  placed. 


B 


B' 


Given  the  similar  polygons  ABCDE  and  A'B'CD'E 
divided  into  triangles  by  the  diagonals  AC,  AD,  and  A'C', 
A' If  drawn  from  the  corresponding  vertices  A and  A' . 

To  prove  that  A ABC,  ACD,  ADE  are  similar  to  the 


& A’B’C',  A'C'D',  A'D'E',  respectively. 

Proof.  1.  ZB  = ZB'.  Art.  321. 

Also  AB  : A'B'  — BC  : B'C'.  Art.  321. 

.*.  AABC  and  A'B'C  are  similar,  Art.  327. 

(if  two  A have  an  Z of  one  = an  Z of  the  other  and  the  including 
sides  proportional,  the  A are  similar). 

2.  Again  /.BCD  = ZB' C'D',  Art.  321. 

( homologous  A of  similar  polygons) . 

Also  ZBCA=  ZB'C'A',  Art.  321. 

(homologous  A of  the  similar  A ABC  and  A'B'C'). 

Subtracting  Z ACD  = Z A'C'D' . Ax.  3. 

But  BG  : B'C'=  CD  : C'D',  Art.  321. 

(homologous  sides  of  similar  polygons  are  proportional) . 

And  BC  : B'C' — AC  : A'C',  Art.  321. 

(being  homologous  sides  of  the  similar  A ABC  and  A'B'C’). 

Hence  AC  : A'C'  — CD  : C'D'.  Ax.  1. 

the  A ACD  and  A'C'D'  are  similar.  Art.  327. 


3.  In  like  manner  it  can  be  shown  that  the  A ADE 
and  A'D’E'  are  similar. 


fi.  E.  D, 


196 


BOOK  III.  PLANE  GEOMETRY 


Proposition  XX.  Theorem  (Converse  of  Prop.  XIX) 


330.  If  two  polygons  are  composed  of  the  same  number 
of  triangles,  similar,  each  to  each,  and  similarly  placed , the 
polygons  are  similar. 


B 


B' 


Given  the  two  polygons  ABCDE  and  A'B'C'D'W,  in 
which  the  A ABC,  ACT),  ABE  are  similar,  respectively,  to 
the  A A'B'C , A'G'B' , A'B'E' , and  are  similarly  placed. 

To  prove  the  polygons  ABCBE  and  A'B'C'B'E'  similar. 

Proof.  ZB  = IB',  Art.  321. 

{being  homologous  A of  similar  A). 

Also  Z BCA  = Z B'C'A'.  (Why?) 

And  ZACD  = ZA'C'D'.  (Why*) 

Adding  equals,  ZBCD  = ZB' C'D' . (Why?) 

In  like  manner  it  may  be  shown  that  Z CDE=  Z C'B'E' , 
ZBAE  = ZB'A'E',  etc. 

Hence  the  polygons  are  mutually  equiangular. 


Also 


A B 
A'B' 


B C 
B'C' 


(homologous  sides  of  similar  A). 


Art.  321. 


. . B C _ A C , CD 

Again  B,G,  A,CI,  and  Gljy 


In  like  manner 


CD  _ D E 
C'D1  D'E' 


= AA 

A'C' 

AE 

A'E'’ 


B C _ C D 
" B'C ' C'D' 

(Why?) 


Hence  the  homologous  sides  of  the  given  polygons  are 
proportional. 

;.  the  polygons  ABCDE  and  A'B'C'D'E'  are  similar. 

(tArt.  321. 

q.  £.  d. 


PROPORTIONAL  LINES 


197 


331.  Note.  It  is  often  convenient  to  write  a series  of  equal 
ratios,  like  those  used  in  the  above  proof,  as  follows: 

A B BC  _ A AC  \ CD  _ (A  I)  \ DE  _ AE 
A'B'~  B'C~  \A'C'J ~ C'D'~  \A'D'J~  D' E’~  A! E1' 
a ratio  which  is  used  merely  to  show  the  equality  of  two  other  ratios 
being  inclosed  in  parenthesis. 

PROPORTIONAL  LINES  AND  NUMERICAL  PROPERTIES 
OF  LINES 

Proposition  XXI.  ■Theorem 

332.  In  any  triangle,  the  bisector  of  an  angle  divides 
the  opposite  side  into  segments  ivhich  are  proportional  to  the 
other  two  sides. 


Given  the  A ABC,  with  the  line  AB  bisecting  the  ABAC, 
and  meeting  BC  at  D. 

To  prove  BC  : BB  = AC  : AB. 

Proof.  Draw  the  line  CF  ||  AB,  and  meeting  AB  pro- 
duced at  F. 


Then 

BC:  BB  — AF  : AB. 

Art.  317. 

But 

Lr—  Zp'. 

Art.  124. 

And 

Zs—  Zp. 

Art.  126. 

Also 

Zp Zp. 

Hyp. 

Z r—  Z s. 

(Why?) 

:.  A ACF  is  isosceles,  and  AC—AF. 

(Why?) 

Substituting  A C for  its  equal,  AF,  in  the  above  proportion, 

BC  : BB  — AC  : AB. 

Ax.  8. 

Ex.  If, 

in  the  above  figure,  AB= 16,  AC ~ 12,  and  BC  — 

Q.  E.  ». 

14,  find  DC 

and  DB. 
[SUG. 

Let  DC=x,  DB= 14  — x,  etc.] 

198 


BOOK  III.  PLANE  GEOMETRY 


333.  A line  divided  internally  is  a line  divided  into  two 
parts  by  a point  taken  between  its  extremities. 

I £ B 

Thus,  the  line  AB  is  divided  internally  at  the  point  P 
into  the  segments  PA  and  PB. 

To  divide  a given  line  internally  in  a given  ratio  (as  2 : 7),  divide  it 
into  a number  of  equal  parte,  equal  to  the  sum  of  the  terms  of  the  ratio 
(as  2 + 7,  or  9 equal  parts). 

334.  A line  divided  externally  is  a line  whose  parts  are 
considered  to  be  the  segments  included  between  a point  on 
the  line  produced  and  the  extremities  of  the  given  line. 

P 7 A B 

Thus,  the  line  AB  is  divided  externally  at  the  point  P 
into  the  two  segments  P'A  and  P'B. 

To  divide  a given  line  externally  ia  a given  ratio  (as  2 : 7),  divide  it 
into  a number  of  equal  parts,  equal  to  the  difference  of  the  terms  of  the 
ratio  (as  7 — 2,  or  5 parts)  and  produce  it  till  the  produced  part  equals 
the  smaller  term  of  the  ratio  times  the  unit  line  found. 

335.  A line  divided  harmonically  is  a line  divided  both 
internally  and  externally  in  the  same  ratio. 

Thus,  if  the  line  AB  is  divided  internally,  so  that  PA  : 
PB  — 1 : 2,  and  externally,  so  that  P'A  : P’B  = 1 : 2,  then 
PA  : PB  = P'A  : P'B,  and  the  line  is  divided  harmonically, 


Ex.  1.  Divide  a given  line  harmonically  in  the  ratio  1 ; 3. 
Ex.  2.  Divide  a given  line  harmonically  in  the  ratio  2 : 5. 


PROPORTIONAL  LINES 


199 


Proposition  XXII.  Theorem 

336.  In  any  triangle,  the  bisector  of  an  exterior  angle 
divides  externally  the  opposite  side  produced  into  segments 
which  are  proportional  to  the  other  two  sides. 


Given  the  A A B G with  its  exterior  /.BAG  bisected  by 
the  line  AD  meeting  the  opposite  side  produced  at  D. 

To  prove  DB  : DG—AB  : AG. 

Proof.  Draw  the  line  BF  ||  AD  and  meeting  AC  at  F. 


Then 

DB  : DG=AF  : AG. 

(Why  1) 

But 

/ r = /p' . 

("Why?) 

And 

/s  — /p. 

(Why?) 

Also 

/p'  — /p. 

(Why?) 

/r  — /s. 

(Why?) 

:.  A ABF  is  isosceles,  and  AB  — AF. 

(Why?) 

Substituting  AB  for  its  equal,  AF,  in  the  above  pro- 
portion, 

DB  : DG—AB  : AC.  Ax.  8. 

Q.  E.  D. 

337.  Cor.  The  lines  bisecting  the  interior  and  exterior 
angles  of  a triangle  at  a given  vertex,  divide  the  opposite 
side  harmonically, 


200 


BOOK  III.  PLANE  GEOMETRY 


Proposition  XXIII.  Theorem 

338.  The  homologous  altitudes  of  tivo  similar  triangles 
have  the  same  ratio  as  any  two  homologous  sides. 


Given  the  similar  A ABC,  A'B'C',  and  BD,  Briy 
any  two  homologous  altitudes  in  these  A. 


To  prove 


BD 

B'D' 


BA 

B'A' 


BC  _AC 
B'C'  A'C>' 


Proof.  In  the  right  A ABD  and  A'B'D1 , 
Zl  = l A', 


Art.  321. 


( '‘ting  homologous  A.  of  the  similar  A ABC  and  A'B'C') . 

A ABD  and  A’B’D'  are  similar,  Art.  324. 
{if  two  rt.  A have  an  acute  Z of  one  = an  acute  Z of  the  other,  the  A 

are  similar). 


BD  BA 


" B'D'  B'A' 

But,  in  the  similar  A ABC  and  A'B'C', 

BA  _BC  ___A  C 
B'A'  B'C'  A' & 

BD  BA  BC  AC 


Hence 


B'D'  B’A'  B’C'  A'C 


Art.  321- 


Art.  321. 

Ax.  1. 
Q.  E.  D. 


Ex.  If,  in  the  above  figure,  AC= 18,  A'C'= 12,  andBI)=15,  findjB'iy. 


PROPORTIONAL  LINES 


201 


Proposition  XXIY.  Theorem 

339.  If  three  or  more  lines  pass  through  the  same  point 
and  intersect  two  parallel  lines , they  intercept  proportional 
segments  on  the  parallel  lines. 


A B c D ' 


D'  C'  B'  A' 


» 

Given  the  transversals  OA,  OB,  00,  OB  intersecting-  the 
parallel  lines  AB  and  A'B'  in  the  points  A,  B,  C,  B and 
Ar,  B' , O',  B ' , respectively. 


To  prove 


AB  BO  OB 
A'B'  B'B  C'B' 


Proof.  A'B'  ||  AB.  Hyp. 

Therefore  the  base  A of  the  A A'B'O,  B'C'O,  etc.,  are 
equal  to  corresponding  base  A of  the  A ABO,  BOO,  etc. 

Art.  126  or  Art.  124. 

Hence  the  A A'B'O,  B'C'O,  etc.,  are  similar  to  the  A 
ABO,  BOO,  etc.,  respectively.  Art.  324. 


AB 

(BO\ 

_B  0 _ 

(CO' 

\GB 

A'B' 

\B'  0) 

B'B 

\C'0 j 

' C'B ' 

Art.  321. 


Q.  E.  D. 


Ex.  If  the  sides  of  a polygon  are  2,  3,  4,  5,  6 feet,  and,  in  a simi- 
lar polygon,  the  side  homologous  to  6 is  9,  find  all  the  other  sides  of 
the  second  polygon. 


202 


BOOK  m.  PLANE  GEOMETRY 


Prop.  XXV.  Theorem  (Converse  of  Prop.  XXIV) 

340.  If  three  or  more  non -parallel  straight  lines  inter- 
cept proportional  segments  on  two  parallels , they  pass  through 
the  same  point  (that  is,  are  concurrent). 


Given  the  transversals  AA' , BB' , CO  intersecting  the 

parallel  lines  AC  and  A'C'  so  that  —rrp;,  = 7777 

A.  x>  x>  0 


To  prove  that  the  lines  AA' , BB',  CO  are  concurrent. 

Proof.  Produce  the  lines  A A'  and  BB'  to  meet  at  some 
point  0. 


Draw  the  line  OC  and  let  it  intersect  the  line  A'C'  at 
some  point  P. 


Then 

But 


AB  _ BC 
A'B'  B'P 
AB  _ B C 
A'B'  B' O' 


Art.  339. 


Hyp  ■ 


Hence 


B'P  =B'C. 


Art.  306 


.*.  point  P coincides  with  point  O. 

.*.  line  CO  coincides  with  line  CP.  Art.  64. 

.*.  the  line  CO,  if  produced,  passes  through  0, 

( for  it  coincides  with  the  line  CP,  which  passes  through  0). 

AA',  BB',  CO  meet  in  Q. 


Q.  E.  D. 


PROPORTIONAL  LINES  203 


Proposition  XXVI.  Theorem 

341.  The  perimeters  of  two  similar  polygons  have  the 
same  ratio  as  any  two  homologous  sides. 


B 


Given  the  two  similar  polygons  ABODE  and  A'B'C'D'E' , 
with  their  perimeters  denoted  by  P and  P'  and  with  AB  and 
A'B'  any  two  homologous  sides. 

To  prove  that  P : P'  = AB  : A'B'. 

Proof.  AB  : A'B'  = BC  : B'C'=OD  : C'D'  = etc.  Art.  321. 

Hence  AB+BC+CD+,  etc.  : A'B'+B'C,+  0'D'+,  etc., 
= AB  : A'B',  Art.  312. 

(in  a series  of  equal  ratios,  the  sum  of  all  the  antecedents  is  to  the  sum  of 
all  the  consequents  as  any  one  antecedent  is  to  its  consequent) . 

Or  P :Pr  = AB  : A'B'. 

Q.  E.  D. 

' 341  (a).  Cor.  In  two  similar  polygons,  any  two  homoU 

ogous  lines  are  to  each  other  as  any  other  two  homologous  lines; 
and  the  perimeters  are  to  each  other  as  any  two  homologous  lines. 


Ex.  If  the  perimeter  of  a given  field  is  210  rods  and  a side  of  this 
field  is  to  a homologous  side  of  a similar  field  as  3 : 2,  find  the  perime- 
ter cf  the  second  field. 


204 


BOOK  III.  PLANE  GEOMETRY 


Proposition  XXVII.  Theorem 

342.  In  a right  triangle, 

I.  The  altitude  to  the  hypotenuse  is  a mean  proportional 
between  the  segments  of  the  hypotenuse ; 

II.  Each  leg  is  a mean  proportional  between  the  hypote- 
nuse and  the  segment  of  the  hypotenuse  adjacent  to  the  given 
leg . 

B 


/j 

a;' N. 

/ X 

rs. 

A F O 


Given  the  right  A ABC  and  BE  the  altitude  upon  thff 
hypotenuse  AC. 

To  prove  I.  AF  : BF=BF  : FC. 

r AC  : AB=AB:  AF 
n'  t AC  : BC=  BC:  FC. 

Proof.  The  Z1  is  common  to  the  rt.  A ABF  and  ABC. 

:.  A ABF  and  ABC  are  similar.  Art.  324. 

Similarly  Z C is  common  to  the  rt.  A BFC  and  ABC, 
and  A BFC  and  ABC  are  similar. 

.*.  A ABF  and  BFC  are  similar,  Art.  325. 

[if  two  A are  similar  to  the  same  A,  they  are  similar  to  each  other). 

Hence,  I.  In  the  A ABF  and  BFC, 


AF  : BF=BF  : FC,  Art.  321. 

( homologous  sides  of  similar  A are  proportional) , 


PEOPOETIONAL  LINES 


205 


II.  In  the  similar  A ABC  and  ABF , 

AC  : AB=AB  : AF.  Art.  321, 

Also,  in  the  similar  A ABC  and  BFC, 

AC  : BC=BC  : FC.  Art.  321. 

Q.  E.  D. 

343.  Cor.  The  perpendicular  to  the  diameter  from  any 
voint  in  the  circumference  of  a circle  is  a mean  pro- 
portional between  the  segments  of  the  di- 
ameter; and  the  chord  joining  the  point  to 
an  extremity  of  the  diameter  is  a mean 
proportional  between  the  diameter  and 
the  segment  of  the  diameter  adjacent  to  the  chord. 

344.  Def.  The  projection  of  a point  upon  a line  is  the 
foot  of  the  perpendicular  drawn  from  the  point  to  the  line. 


i i I 

L A7  ~B’  AT  P V Q 

Thus,  if  AA'  be  perpendicular  to  LM , A ' is  the  projec- 
tion of  the  point  A on  the  line  LM. 

345.  The  projection  of  a line  upon  another  given  line  is 
that  part  of  the  second  line  which  is  included  between  per- 
pendiculars drawn  from  the  extremities  of  the  first  line 
upon  the  second  line.  Thus,  the  projection  of  AB  on  LM 
is  A'B'\  of  CD  on  PQ,  is  C'D. 

Ex.  1.  If  the  segments  of  the  hypotenuse  of  a right  triangle  are  3 
and  12,  find  the  altitude  on  the  hypotenuse;  also  find  the  legs  of  the 
triangle. 

Ex.  2.  If,  in  the  figure  of  Art.  343,  the  diameter  is  20  and  the  longer 
chord  16,  find  tha  segment  of  the  diameter  adjacent  to  the  chord. 


206 


BOOK  III.  PLANE  GEOMETRY 


Proposition  XXVIII.  Theorem 

346.  In  a right  triangle,  the  square  of  the  hypotenuse  is 
equal  to  the  sum  of  the  squares  of  the  legs. 


Given  the  right  triangle  ABC  with  AC  the  hypotenuse. 
To  prove  AC2=Zb2  + PC2. 

Proof.  Draw  the  line  BF  A AC.  Then 
AC  : AB  = AB  : AF  :.AC  X AF=  AB2.  Arts.  342,  302. 
Also  AC  : BC=BC  : FC  :.  AC  X FC=BC2.  (Why?) 
Adding  equals,  AC  ( AF  + FC)  =AB~  -f  BC2.  Ax.  2. 
Or  AC2  — AB~  BC 2 Axs.  6,  8. 

Q.  E.  D. 

347.  Cor.  1.  In  a right  triangle,  the  square  of  either 
leg  is  equal  to  the  square  of  the  hypotenuse  minus  the  square 
of  the  other  leg. 


348.  Cor.  2.  In  the  square  ABCB,  the  diagonal  di 
vides  the  square  into  two  right  triangles. 

Hence  AC2  = AB2  + PC2  = 2 AP2. 

AC=ABV2,  orj|  = 1^- 

Hence  the  diagonal  and  the  side  of  a 
square  are  incommensurable . 

Ex.  1.  If  the  legs  of  a rt.  A are  li  in.  and  2 in.,  find  the  hypotenuse 
Ex.  2.  In  the  figure  on  p.  204,  show  that  AB~  : £C~=AF  : FC, 


NUMERICAL  PROPERTIES 


207 


Proposition  XXIX.  Theorem 

349.  In  any  oblique  triangle , the  square  of  a side  oppo- 
site an  acute  angle  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides , diminished  by  twice  the  product  of  one  of 
those  sides  by  the  projection  of  the  other  side  upon  it. 


Given  acute  Z (7  in  A ABC,  and  DC  the  projection  of 
the  side  BC  on  the  side  AC. 

To  prove  Zb2  = J^2+7o2  — 2 AC  X DC. 

Proof.  If  D falls  on  AC  (Fig.  1),  AD  = AC  — DC. 

If  D falls  on  AC  produced  (Fig.  2),  AD  = DC — AC. 

In  either  case,  AD2  = AC2  -+-  DC 2 — 2 AC  X DC.  Ax.  4. 

Adding  BD“  to  each  of  these  equals, 

AD 2 + BD2  = AC 2 + Ixf  + BD2—  2 AC  X DC.  Ax.  2. 

But,  in  the  rt.  A ABD,  AD2  + BD2  = AB2,  Art.  346. 

and,  in  the  rt.  A DBC,  DC'  + BD2—BC2.  (Why?) 

Substituting  these  values  in  the  above  equality, 

AB2  = BC2  + AC2  — 2 ACX  DC.  Ax.  8. 

Q.  E.  D. 

fix.  1.  A line  10  in.  long  makes  an  angle  of  45°  with  a second 
line;  find  the  projection  of  the  first  line  on  the  second. 

Ex.  2.  Find  the  same,  if  the  angle  is  60°. 

Ex.  3.  If  the  side  of  an  equilateral  triangle  is  a,  find  its  projeo» 
tion  on  the  base. 

Ex.  4.  If,  in  Pig.  £C=10,  AC=  12,  and  £C=  60°,  find  AB. 


208 


EOOK  in.  PLANE  GEOMETRY 


Proposition  XXX.  Theorem 

350.  In  any  obtuse  triangle , the  square  of  the  side  oppo- 
site an  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides,  increased  by  twice  the  product  of  one  of  the 
sides  by  the  projection  of  the  other  side  upon  that  side. 


Given  the  obtuse  Z.ACB  in  tb.e  A ABC,  and  CD  the 
projection  of  BC  on  AC  produced. 

To  prove  AB2 = AC2+BC2+  2 AC  X CD. 

Proof.  AD=ACA  CD.  Ax.  6. 

Zd2=~AC2A  CD2 +2  AC  X CD.  Ax.  4. 

Adding  BIX  to  each  of  these  equals, 

Id2 a bd2=  Tc2a  op2 a bd2a  2 acx  cd.  ax.  2. 

But,  in  the  rt.  A ABD,  AD2  A BD~=  AB2.  {Why  ?) 

And,  in  the  rt.  A CBD,  CD2  A BD~=  BC2.  (Why?) 

Substituting  these  values  in  the  above  equality, 

Ip2  - ~AC2A  BC2A  2 ACX  CD.  Ax.  8. 

Q.  E.  D. 

351.  Cor.  If  the  square  on  one  side  of  a triangle  equals 
the  sum  of  the  squares  on  the  other  two  sides,  the  angle  oppo- 
site the  first  side  is  a right  angle;  for  it  cannot  be  acute 
(Art.  349),  or  obtuse  (Art.  350). 

Ex.  1.  If,  in  the  above  figure,  BC=  10,  AC= 2,  and  ZBCA  = 120°, 
find  AB. 


Ex.  2.  If  AB=20,  EC=14  and  AC- 12,  find  CD. 


NUMERICAL  PROPERTIES 


209 


Proposition  XXXI.  Theorem 

352.  If,  in  any  triangle , a median  be  drawn  to  one  side, 

I.  The  sum  of  the  squares  of  the  other  two  sides  is  equal 
to  twice  the  square  of  half  the  given  side,  increased  by  tivice 

' the  square  of  the  median  upon  that  side  ; and 

II.  The  difference  of  the  squares  of  the  other  two  sides  is 
equal  to  twice  the  product  of  the  given  side  by  the  projection 
of  the  median  upon  that  side. 


A 


Given  the  A ABC,  AB>AG,  AM  the  median  upon  BG, 
and  MF  the  projection  of  AM  on  BG. 

To  prove  I.  Zb2+  ~AG2 —2  ~BM2+  2 ~AM2. 

II.  AB2—  1g2  = 2 BGXMF. 

Proof.  In  the  A BMA  and  AMC,  BM=MC,  AM=AM, 
and  AB>AC.  Hyp. 

/.  ZAMB  is  greater  than  Z AMC.  Art.  108. 

Z AMB  is  obtuse  ( for  it  is  greater  than  half  a straight  Z ). 

In  obtuse  A ABM,  AB2=BM2+AM2+  2 BM  X MF. 

Art.  350. 

In  acute  A ACM,  AC2  = MG2 + AM2—  2 MG  X MF. 

Art.  349. 

Adding,  Zb2+  AC2  = 2 BM2+  2 AM2,  Ax.  2. 

( for  MC=BM). 

Subtracting,  AB2 — AC2=2BCXMF,  Axs.  2,  6. 

(/or  BM+MC=BC). 

Q.  £.  D. 


N 


210 


BOOK  III.  PLANE  GEOMETRY 


353.  Formula  for  median  of  a triangle  in  terms  of  its 
sides.  In  tlie  Fig.,  p.  209,  denoting  AB  by  c,  AC  by  h, 
BC  by  a,  and  AM  by  m,  by  Art.  352,  h2-\-c2  = 2m2 

+ 2 ^ j--Q , whence  m = i\/Z(h2  + c2)  — a2. 

Proposition  XXXII.  Theorem 

354.  If  tivo  chords  in  a circle  intersect,  the  product  of 
the  segments  of  one  chord  is  equal  to  the  product  of  the  seg- 
ments of  the  other  chord. 

D 


Given  the  O ADBC  with  the  chords  AB  and  CD  inter- 


secting at  the  point  F. 

To  prove  AFX  FB—  CF  X FD. 

Proof.  Draw  AD  and  CB. 

Then,  in  the  A AFD  and  CFB,  Zi=  Z C,  Art.  258. 
( each  being  measured  by  i arc  DB ) . 

Also  Z J>=  Z B.  (Why?) 

Hence  the  A AFD  and  CFB  are  similar.  (Why  ?) 

/.  AF  : CF=FD  : FB.  (Why?) 

And  AF  X FB  = CF  X FD.  (Why  ?) 


Q.  E.  D. 

355/  Cor.  1.  If  through  a fixed  point  within  a circle  a 
chord  he  drawn,  the  product  of  the  segments  of  the  chord  is 
constant,  in  whatever  direction  the  chord  he  drawn. 


proportional  lines 


211 


356.  Def.  Four  directly  proportional  quantities  are  four 
quantities  in  proportion  in  such  a way  that  both  the  ante- 
cedents belong  to  one  figure  and  both  the  consequents  to 
another  figure. 

Four  reciprocally  proportional  quantities  are  four  quanti- 
ties in  proportion  in  such  a way  that  the  means  belong  to 
one  figure  and  the  extremes  to  another  figure. 

357.  Cor.  2.  The  segments  of  two  chords  intersecting 
in  a circle  are  reciprocally  proportional  (the  segments  be- 
ing considered  as  parts  of  the  chords,  not  of  the  A). 

Proposition  XXXIII.  Theorem 

358.  If,  from  a given  point,  a secant  and  a tangent  he 
drawn  to  a circle,  the  tangent  is  the  mean  proportional  be- 
tween the  whole  secant  and  its  external  segment. 


A 


Given  AB,  a tangent,  and  AC,  a secant,  to  the  circle 
BCF,  and  AF  the  external  segment  of  the  secant. 

To  prove  AC  : AB—AB  : AF. 

Proof.  Draw  the  chords  BC  and  BF. 

Then,  in  the  A ABC  and  ABF,  Z A — Z A.  Ident. 

ZC-/.ABF.  Arts.  258,  264. 

the  A ABC  and  ABF  are  similar.  (Why  ?) 

.-.  AC  : AB  — AB  : AF.  (Why  ?) 

Q.  E.  D. 


212 


BOOK  III.  PLANE  GEOMETEY 


359.  Cor.  1.  If,  from  a given  point , a tangent  and  a 
secant  be  drawn  to  a circle,  the  product  of  the  ivhole  secant 
and  its  external  segment  is  equal  to  the  square  of  the  tangent. 

360.  Cor.  2.  If,  from  a given  point  without  a circle,  a 
secant  be  drawn,  the  product  of  the  secant  and  its  external 
segment  is  constant,  in  whatever  direction  the  secant  be  drawn. 

For  the  product  of  each  secant  and  its  external  seg 
ment  equals  the  square  of  the  tangent,  which  is  constant. 

361.  Cor.  3.  If  two  secants  be  draivn  from  an  exter- 
nal point  to  a circle,  the  whole  secants  and  their  external 
segments  are  reciprocally  proportional. 


Proposition  XXXIV.  Theorem 


362.  The  square  of  the  bisector  of  an  angle  of  a triangle 
is  equal  to  the  product  of  the  sides  forming  the  angle,  dimin- 
ished by  the  product  of  the  segments  of  the  third  side  formed 
by  the  bisector. 


H 


Given  the  A ABC,  CF  (or  t)  the  bisector  of  Z ACB , 
and  m and  n the  segments  of  AB  formed  by  CF. 

To  prove  t~  = ab  — mn . 

Circumscribe  a Q about  the  A ABC . 


NUMERICAL  PROPERTIES 


213 


Proof.  Produce  CF  to  meet  the  circumference  at  R,  and 


draw  the  chord  BR. 

Then,  in  the  A ACF  and  CRB,  Z A = ZH.  (Why?) 
And  ZACF=  ZRCB.  (Why?, 

Hence  the  A ACF  and  CRB  are  similar.  (Why  ?) 

.'.  b : t = X-\-t  : a.  (Why?) 

,\  t ( x-\-t)—ab . (Why  s ) 

Or  tx  + t2  = ab. 

Subtracting  tx,  t2=ab — tx.  Ax.  3. 

But  tx=mn.  Art.  354. 

Substituting  mn  for  tx, 

t2=ab — mn.  Ax.  8. 


Q.  E.  D. 


363.  Formula  for  bisector  of  an  angle  of  a triangle. 

m : n — b : a (Art.  332)  m + n :m  = b + a : b (Art.  309) , 

be 

or  c : m = a-\-  b : b (Ax.  8)  m — — — - (Art.  302,  Ax.  5). 

a -j-b 

GC 

In  like  manner  n— — — ■ Substituting  for  m and  n, 
a + b 

abc 2 _ ab(a  + b + c)  (a  + b — c) 

t'~ab~  (a  + 6)2  ~ (o  + bf- 

Let  a 4-  b + c = 2s;  then  a + b — c = 2s  — 2c  = 2(s  — ■ c 

2 / — 

Then  t -r—;Vabs{s  — c). 

a + b 


Ex.  1.  On  the  figure,  p.  210,  let  AF=  10,  FB= 4,  and  FC= 8. 
Find  FD. 

Ex.  2.  On  the  figure,  p.  211,  let  AC=  16  and  AB= 12.  Find  AF. 

Ex.  3.  On  the  figure,  r 212,  let  AC=16,  CB= 12,  and  AB= 14. 
Find  CF. 


214 


BOOK  III.  P.oaNE  GEOMETRY 


Proposition  XXXV.  Theorem 

364.  In  any  triangle , the  product  of  any  two  sides  is 
equal  to  the  product  of  the  diameter  of  the  circumscribed  cir- 
cle by  the  altitude  upon  the  third  side. 


D 


Given  the  A ABC , ABCD  a circumscribed  O,  BD  the 
diameter  of  this  O,  and  BF  the  altitude  upon  AC. 


To  prove  AB  X BC=  BD  X BF. 

Proof.  Draw  the  chord  DC. 

• Then,  in  the  A ABF  and  DBG,  I.A  — /.D.  (Why  t) 
Z DCB  is  a rt.  Z . (Why?) 

.*.  A ABF  and  DBC  are  similar.  (Why?) 

.*.  AB  : BD  — BF  : BC.  (Why!) 

.’.  AB  X BC=BD  X BF.  (Why?) 


Q.  E.  D. 

365.  Cor.  The  diameter  of  a circle  circumscribed  about 
a triangle  equals  the  product  of  two  sides  of  the  triangle 
divided  by  the  altitude  upon  the  third  side. 


Ex.  In  the  above  figure,  if  AB  = 8,  BC  = 6,  and  BF  = 4|,  find  the 
radius  of  the  circumscribed  circle. 


CONSTRUCTION  PROBLEMS 


215 


CONSTRUCTION  PROBLEMS 

Proposition  XXXVI.  Problem 

366.  To  construct  a fourth  proportional  to  three  given 
lines. 


Given  the  lines  m,  n,  p. 

To  construct  a fourth  proportional  to  m,  n,  and  p. 
Construction.  Take  any  two  lines,  AP  and  making 
any  convenient  /.A. 

On  AP  take  AB  = m,  and  BC=n. 

On  AQ  take  AB=p. 

Draw  PJ>.* 

Through  the  point  C draw  CP  1 1 BB,  and  meeting  AQ 
at  R.  Art.  279. 

Then  BB  is  the  fourth  proportional  required. 

Proof.  AB  : BC—AB  : BP.  Art.  317. 

Or  m : n=p  : BB.  Ax.  8. 

Q.  E.  F. 

Proposition  XXXVII.  Problem 

367.  To  construct  a third  proportional  to  two  given  lines 

Let  the  pupil  supply  the  construction  and  proof. 

[Sug.  Use  the  method  of  Art.  366,  making  p = n.~\ 

Ex.  1.  Construct  a fourth  proportional  to  three  lines,  t,  1 and 
H in.  long. 

Ex.  2.  Construct  a third  proportional  to  two  lines,  2 and  li  in  long. 

* Prom  now  on,  references  to  the  postulates  will  be  omitted  where  these  are  not 
necessary  to  complete  some  other  reference. 


216 


BOOK  HI.  PLANE  GEOMETRY 


Proposition  XXXVIII.  Problem 


Given  the  lines  m and  n. 

To  construct  a mean  proportional  between  m and  n. 

Construction.  On  the  line  AP  take  AB  = m,  and  BC—n. 

On  AG  as  a diameter  construct  a semi -circumference. 

Art.  275,  Post.  3. 

At  B erect  a A to  AC  meeting  the  semi -circumference 
At  E.  Art.  274. 

Then  BE  is  the  mean  proportional  required. 

Proof.  AB  : BE  = BE  : BC,  Art.  343. 

( the  A.  to  the  diameter  from  any  point  in  the  circumference  of  a circle  is 
a mean  proportional  between  the  segments  of  the  diameter). 

Substituting  for  AB  and  BC  their  values  m and  n, 

m '■  BE  = BE  : n.  Ax.  8. 

Q.  E.  F. 

Ex.  1.  Construct  the  third  proportional  to  two  lines,  1 and  H in. 
long. 

Ex.  2.  Construct  a mean  proportional  between  two  lines,  1 and  2 
in.  long. 

Ex.  3.  Taking  any  line  as  1,  construct  y/2. 


CONSTRUCTION  PROBLEMS 


217 


Proposition  XXXIX.  Problem 


369.  To  divide  a given  straight  line  into  parts  propor- 
tional to  a number  of  given  lines. 


n ' 


D" 


■VB 


P 

Given  the  straight  lines  AB,  m,  n and  p. 

To  divide  AB  into  parts  proportional  to  m,  n and  p. 
Construction.  Draw  the  line  AP,  making  any  convenient 
angle  with  AB. 

On  APmark  off  A (7=  m,  CB  — n , andDP=p.  DrawPP. 
Through  the  points  C and  D draw  lines  ||  BF,  and  meet- 
ing AB  at  B and  S.  Art.  279. 

Then  AB,  BS  and  SB  are  the  segments  required. 
AB==BS=SB 
A G CD  BF' 

[if  two  lines  are  cut  by  a number  of  parallels , the  corresponding  segments 
are  proportional) . 

For  AC,  CD  and  BF  substitute  their  equals  m,  n and  p. 
AB  BS  SB 

— — Ax.  8. 

m n p 

Q.  E.  F. 


Proof. 


Then 


Art.  319. 


370.  Def.  A straight  line  divided  in  extreme  and  mean 

ratio  is  a straight  line  divided  into  two  segments  such  that 
one  of  the  segments  is  a mean  proportional  between  the  whole 
line  and  the  other  segment. 


Ex.  Divide  a given  line  into  parts  proportional  to  2,  3 and  4. 


218 


BOOK  III.  PLANE  GEOMETRY 


Proposition  XL.  Problem 

371.  To  divide  a given  straight  line  in  extreme  and 
mean  ratio. 


Given  the  line  AB. 

To  divide  AB  in  extreme  and  mean  ratio. 

Construction.  At  one  end  of  the  given  line,  as  B,  con 
struct  a -L  OB  equal  to  \ AB.  Arts.  274,  275, 

From  0 as  a center  and  with  OB  as  a radius,  describe 
a circumference. 

Draw  AO  meeting  the  circumference  at  C,  and  produce 
it  to  meet  the  circumference  again  at  F. 

On  AB  mai-k  off  AP  equal  to  A C;  on  BA  produced 
take  AQ=AF. 

Then  AB  is  divided  in  extreme  and  mean  ratio  inter- 
nally at  P,  and  externally  at  Q. 

Proof.  1.  AF  : AB  = AB  : AO.  Art.  358. 

.*.  AF — AB  : AB  = AB  — AG:  AO.  (Why?) 

But  CF=20B  = AB,  and  AC=AP. 

Hence  AF — AB  — AP,  and  AB  — AC—PB. 

.-.  AP:  AB  = PB  : AP,  or  AB  : AP  = AP  : PB. 


Ax.  8.  Art.  308. 

2.  AF  : AB  = AB  : AC.  (WhyT) 

A AF+AB:  AF=AB  + AC:  AB.  (WhyT) 
But  AF+  AB=QB,  and  AB  -\-AC—AF=  QA. 

QB  : QA  =■  QA  : AB.  Ax.  8. 

£.  I. 


CONSTRUCTION  PROBLEMS 


219 


Proposition  XLI.  Problem 

372.  Upon  a given  straight  line,  to  construct  a polygon 
similar  to  a given  polygon,  and  similarly  placed. 


Given  the  polygon  ABODE  and  the  line  A'B' . 

To  construct  on  A'B'  a polygon  similar  to  ABODE  and 
similarly  placed. 

Construction.  In  the  given  polygon,  draw  the  diagonals 
AC  and  AD,  dividing  the  polygon  into  triangles. 

At  B'r on  the  line  A'B' , construct  Z A'B' O'  equal  to  Z B\ 
and  at  A' construct  LB' A' O'  equal  to  LB  AG.  Art.  278. 

Produce  the  lines  B’G'  and  A'C'  to  meet  at  G' . 

In  like  manner,  on  A'C'  construct  A A' CD' , equiangular 
with  A AGD  and  similarly  placed;  an!  on  A'D'  construct 
the  A A'D'E' , equiangular  with  A ADE  and  similarly 
"placed. 

Then  A'B'C'D'E'  is  the  polygon  required. 

Proof.  The  A A'B'C',  A' C'D',  etc.,  are  similar  to  the 
A ABC,  AGD,  etc.,  respectively.  Art.  324. 

Hence  the  polygons  A'B'C'D'E'  and  ABODE  are 
similar.  Art.  330. 


D 


C 


Q.  E.  F. 


220 


BOOK  III.  PLANE  GEOMETRY 


EXERCISES.  CROUP  30 

SIMILAR  TRIANGLES 

Let  the  pupil  make  a list  of  all  the  conditions  that  make 
ttvo  triangles  similar  (see  Arts.  323,  324,  325,  etc.). 

Ex.  1.  Given  AD  J.  BC,  and  BF  JL  AC;  prove 
A ADC  and  BFC  similar. 

Ex.  2.  In  the  same  figure,  prove  the  A AOF 
and  BFC  similar.  What  other  triangle  on  this 
figure  is  similar  to  A BFC  ? 

Ex.  3.  Two  isosceles  triangles  are  similar  if 
their  vertex  angles  are  equal. 

Ex.  4.  Two  isosceles  triangles  are  similar  if  a 
equals  a base  angle  of  the  other. 

Ex.  5.  Given  arc  AC=  arc  BC;  prove  A 
APC  and  AFC  similar. 

Ex.  6.  Prove  that  the  diagonals  and  bases 
of  a trapezoid  together  form  a pair  of  similar 
triangles. 

Ex.  7.  AB  is  the  diameter  of  a circle,  BD  is 
a tangent,  and  AD  intersects  the  circumference  at  E.  Prove  the  tri- 
angles ABE  and  ADB  similar. 

Ex.  8.  BC  is  a chord  in  a circle,  AQ  is  the  diameter  perpendicular 
to  BC  and  meeting  it  at  N;  AP  is  any  chord  intersecting  BC  in  AC. 
Prove  the  A AMN  and  APQ  similar. 

Ex.  9.  The  triangle  ABC  is  inscribed  in  a circle;  the  bisector  of 
the  angle  A meets  BC  in  D and  the  circumference  in  P.  Prove  th- 
triangles  BAD  and  APC  similar. 

Ex.  10.  A pair  of  homologous  medians  divide  two  similar  triangles 
into  triangles  which  are  similar  each  to  each. 

Ex.  11.  Two  rectangles  are  similar  if  two  adjacent  sides  of  one 
are  proportional  to  the  homologous  sides  of  the  other. 

Ex.  12.  Two  circles  intersect  in  the  points  A and  B.  AC  and  AD 
are  each  a tangent  in  one  circle  and  a chord  in  the  other.  Prove  the 
A ABC  and  ABD  similar. 

[Sug.  Prove  Z.BAD=  Z ACB,  etc. 2 


C D 
base  angle  of  one 


EXERCISES.  PROPORTIONAL  LINES  221 

373.  Proof  that  lines  are  proportional.  In  order  to 
prove  that  certain  lines  are  proportional,  or  have  propor- 
tional relations,  it  is  usually  best  to  show  that  the  given 
lines  are  homologous  sides  of  similar  triangles. 

Sometimes,  however,  other  methods  of  proof  are  used 
(as  the  theorems  of  Arts.  354  and  358) ; but  these,  if  in- 
vestigated, are  usually  found  to  be  the  method  of  similar 
triangles  in  disguise. 

EXERCISES.  CROUP  31 

. PROPORTIONAL  LINES 

Ex.  1.  On  the  figure  of  Ex.  1,  p.  220,  prove  ADX  BC=BFX  AC, 
and  BCX  OD=BOX  FC. 

Ex.  2.  On  the  figure  of  Ex.  5,  p.  220,  prove  CP  ■ CA  = CA  : CF. 
(Hence  as  P moves  the  product  of  what  two  lines  is  constant  ? ) 

Ex.  3.  The  diagonals  of  a trapezoid  divide  each  other  into  pro- 
portional segments. 

Ex.  4.  In  the  isosceles  triangle  ABC,  AB—AC,  on  the  side  AB, 
the  point  Pis  taken  so  that  PC  equals  the  base.  Prove  ABX  PB=BC2. 

Ex.  5.  In  a triangle  the  median  to  the  base  bisects  all  lines 
parallel  to  the  base  and  terminated  by  the  sides. 

Ex.  6.  If  PQ  is  any  line  through  F,  the  midpoint  of  the  line  AB, 
and  AP  and  BQ  are  perpendicular  to  PQ,  show  that  the  ratio  PF:  FQ 
is  constant. 

Ex.  7.  The  triangle  ABC  is  inscribed  in  a circle.  Pis  the  midpoint 
of  the  arc  AC,  and  BF  intersects  the  line  AC  in  E.  Prove  AB  : BC  = 
AE:  EC. 

[Sue.  Use  Art.  332.] 

Ex.  8.  If  two  circles  intersect,  the  common  chord,  if  produced, 
oisects  the  common  tangent. 

[Sug.  Use  Art.  358.] 

Ex.  9.  If  two  circles  intersect,  tangents  drawn  to  the  two  circles 
from  any  point  in  the  common  chord  produced  are  equal. 


222 


BOOK  III.  PLANE  GEOMETRY 


Ex.  10.  Given  AD,  PT  and  BC  ||  ; 
prove  PQ  = RT. 

pr>  PT 

[Sug.  Show  that  r~=  by  show- 
1)0 

ing  them  equal  to°a  common  ratio.] 


Ex.  11.  Lines  are  drawn  from  a within  the  triangle,  to  the 

vertices  of  the  triangle  ABC.  From  B'  any  point  in  OB,  B'A'  is 
drawn  parallel  to  BA  and  meeting  OA  in  A',  and  B'C’  is  drawn  parallel 
to  BC  and  meeting  OC  in  C' . Prove  AlB'  : AB=B'C'  : BC,  and  the 
triangles  ABC  and  J.'B'C'  similar. 

Ex.  12.  Given  AB CD  a OJ  , and 
P any  point  in  BC  produced;  prove 

AR^BQXBP- 

[Sug.  Compare  the  similar  A 
ABR  and  RQD  ; also  the  similar  A 
ARD  and  EBP.] 


EXERCISES.  CROUP  32 

NUMERICAL  PROPERTIES  OP  LINES 
Ex.  1.  If  AD  is  the  altitude  of  the  triangle  ABC,  AB 2 — AC2= 

bd2—dc2. 

Ex.  2.  If  the  diagonals  of  a quadrilateral  are  perpendicular  to 
each  other,  the  sum  of  the  squares  of  one  pair  of  opposite  sides 
equals  the  sum  of  the  squares  of  the  other  pair  of  sides. 

Ex.  3.  The  square  of  the  altitude  of  an  equilateral  triangle  is 
three-fourths  the  square  of  one  side. 

Ex.  4.  If  AB  is  the  hypotenuse  of  a right  triangle,  and  the  leg  BC 

is  bisected  at  K,  AB 2 — A K~  = 3 OK2.  > 

\ 

Ex.  5.  PQ  is  a line  parallel  to  the  hypotenuse  AB  of  a right  triangle 
ABC,  and  meeting  AC  in  P and  BC  in  Q.  Prove  AQ  ~ + BP~=AB~-]-PQ". 

Ex.  6.  In  the  right  triangle  ABC,  BE  and  CF  bisect  the  legs  AC 
and  AB  in  the  points  E and  F.  Prove  4RE"-|-4CE"  = 5ISC". 


EXERCISES.  AUXILIARY  LINES 


223 


EXERCISES.  CROUP  33 


AUXILIARY  LINES 


Ex.  1.  Given  ABCD  a rectangle;  prove 
PA2  + PC2= PB2  + PD2. 

Ex.  2.  The  common  tangent  of  two  cir- 
cles divides  the  line  of  centers  into  segments 
which  have  the  same  ratio  as  the  diameters 
of  the  circles. 


[Sug.  Dra.w  radii  to  the  points  of  contact.] 


Ex.  3.  AB  and  AC  are  the  legs  of  an  isosceles  triangle  and  BF  is 
akt  altitude.  Prove  2 ACX  FC=BC2. 

Ex.  4.  Given  the  chords  AB  and  CD  perpen- 
dicular to  each  other  and  intersecting  at  0 ; prove 
OA2  + OR2  + OC2  + OD2=(diameter)2. 

[Sug.  Draw  the  diameter  BE  and  the  chords 
AC,  BD,  DE.  Prove  AC=ED,  etc.] 

D 

Ex.  5.  ABC  is  an  inscribed  isosceles  triangle 
of  which  AB  and  AC  are  the  legs.  AD  is  a chord  meeting  BC  in  E. 
Prove  AB2=ADXAE. 


Ex.  6.  If  C is  the  vertex  of  an  isosceles  triangle  ABC,  and  D is  a 
point  in  the  base  produced,  then  CD2=CB2  + ADX  BD. 

Ex.  7.  Two  circles  touch  at  the  point  T.  PTP'  and  QTQ'  are  lines 
drawn  meeting  the  circumferences  in  P,  Q and  P' , Q'  respectively. 
Prove  the  triangles  PTQ  and  P'TQ'  similar. 

[Sug.  Draw  the  common  tangent  at  T.] 

Ex.  8.  Two  circles  touch  at  the  point  T,  through  T three  lines  are 
drawn  meeting  the  circumferences  in  P,  Q,  B and  P' , O',  B',  respec- 
tively. Prove  the  triangles  PQB  and  P'Q’PJ  similar. 

Ex.  9.  If  A is  the  midpoint  of  CD,  an  arc  of  a circle,  and  AP  is 
any  chord  intersecting  the  chord  CD  in  Q,  prove  that  APXAQ  is  a 

constant. 


224 


BOOK  III.  PLANE  GEOMETRY 


Ex.  10.  In  an  inscribed  quadrilateral,  the 
product  of  the  diagonals  is  equal  to  the  sum  of 
the  products  of  the  opposite  sides. 

[Sue.  Draw  BF  so  that  Z CBF=  Z ABB  and 
use  similar  triangles.] 


Ex.  11.  The  sum  of  the  squares  of  the 
sides  of  any  quadrilateral  is  equal  to  the  sum 
of  the  squares  of  the  diagonals,  plus  four  times 
the  square  of  the  line  joining  the  midpoints  of 
the  diagonals. 


EXERCISES.  CROUP  34 


INDIRECT  DEMONSTRATIONS 


Ex.  1.  If  the  sum  of  the  squares  on  two  sides  of  a triangle  is 
greater  than  the  square  on  the  third  side,  the  angle  included  by  the 
two  given  sides  is  an  acute  angle. 

Ex.  2.  If  D is  a point  in  the  side  AC  of  the  triangle  ABC,  and 
AD  : DC=AB  : BC,  then  DB  bisects  angle  ABC. 

Ex.  3.  A given  straight  line  can  be  divided  in  a given  ratio  at  but 
one  point. 

Ex.  4.  If  the  sides  of  two  triangles  are  par- 
allel, each  to  each,  and  a straight  line  be  passed 
through  each  pair  of  homologous  vertices,  these 
lines,  if  produced,  will  meet  in  a common  point. 

Ex.  5.  If  each  of  three  circles  intersects  the 
other  two,  the  three  common  chords  intersect  in 
one  point. 

EXERCISES.  CROUP  35 

THEOREMS  PROVED  BY  VARIOUS  METHODS 


Ex.  1.  In  the  figure  on  p.  204,  show  that  Hi?X  BF=BCy^  AF. 

Ex.  2.  In  the  same  figure,  if  FC=3AF,  show  that  Hi)2  : i>’G'2  = l : 3. 
Ex.  3.  AB  is  the  diameter  of  a circle  and  PB  is  a tangent.  If  AP 
meets  the  circumference  in  the  point  Q,  prove  that  AP  X AQ=AB2, 


MISCELLANEOUS  EXERCISES.  THEOREMS 


225 


Ex.  4.  If  the  line  bisecting  the  parallel  sides  of  a trapezoid  be 
produced,  it  meets  the  legs  produced  in  a common  point. 

[Sug.  See  Art.  340.] 

1 Ex.  5.  In  similar  triangles,  homologous  medians  have  the  same 
ratio  as  homologous  sides. 

Ex.  6.  A diameter  AB  is  produced  to  the  point  6';  CF  is  perpen- 
dicular to  AC;  PB  produced  meets  the  circumference  at  Q.  Prove, 
the  triangles  AQB  and  PCB  similar. 

Ex.  7.  If  PA  and  PB  are  chords  in  a circle,  and  CD  is  a line  par- 
allel to  the  tangent  at  P and  meeting  PA  and 
PB  at  C and  D,  the  triangles  PAB  and  PCD  are  D 
similar. 

Ex.  8.  Given  AB  a diameter  and  AD  and  BC 
tangents,  A C and  DB  intersecting  at  any  point  F 
on  the  circumference;  prove  AB  a mean  propor- 
tional between  the  sides  AD  and  BC. 

Ex.  9.  In  any  isosceles  triangle,  the  square  ^ 
of  one  of  the  legs  equals  the  square  on  a line 
drawn  from  the  vertex  to  any  point  of  the  base 
plus  the  product  of  the  segments  of  the  base. 

Ex.  10.  A line  drawn  through  the  intersection  of  the  diagonals  of 
a trapezoid  parallel  to  the  bases  and  terminated  by  the  legs  is  bisected 
by  the  diagonals. 

| Sug.  See  Ex.  10,  p.  222.] 

Ex.  1 1.  If  a chord  is  bisected  by  another  chord,  each  segment  of 
the  first  chord  is  a mean  proportional  between  the  segments  of  the 
second  chord. 

Ex.  12.  In  a parallelogram  the  sum  of  the  squares  of  the  sides 
equals  the  sum  of  the  squares  of  the  diagonals. 

Ex.  13.  If  two  circles  are  tangent  externally,  and  a line  is  drawn 
through  the  point  of  contact  terminated  by  the  circumferences,  the 
chords  intercepted  in  the  two  circles  are  to  each  other  as  the  radii. 

Ex.  14.  Three  times  the  sum  of  the  squares  of  the  sides  of  a tri- 
angle equals  four  times  cv>m  of  the  squares  of  the  medians. 


O 


226 


BOOK  III.  PLANE  GEOMETRY 


Ex.  15.  Find  the  locus  of  the  midpoints  of 
lines  in  a triangle  parallel  to  the  base  and  ter- 
minated by  the  sides. 

Ex.  16.  Given  AB  the  diameter,  AP,  PQB, 
BB,  tangents  ; prove  PQ  X QB  a constant  (=  ra- 
dius squared). 

Ex.  17.  O is  the  center  of  a circle  and  A is 
any  point  within  the  circle ; OA  is  produced  to  B, 
so  that  OA~XOB  equals  the  radius  squared. 

If  P is  any  point  in  the  circumference,  the 
angles  OP A and  OBP  are  equal. 

[Sug.  Use  Art.  327.] 


Ex.  18.  Given  AF=FB,  and  CR  ||  AB 
prove  HP  : FP=RE  : FK. 

[Sue.  RP  : FP=CR  : FB,  etc.] 

Ex.  19.  If  from  any  point  P within  the  triangle  ABC  the  perpen- 
diculars PQ,  PR,  PT  are  drawn  to  the  sides  All,  AC,  BC,  respectively, 
then  ~AQ2  + BIs  + lRC2=Ali 2 + ~BQ2  + CT2. 


EXERCISES.  CROUP  36 

PROBLEMS 

Given  three  lines  a,  b,  c, 

Ex.  1.  Construct  %=  — : also 

c 2c 

Ex.  2.  Construct  x=\/  a 2 — b 2,  i.  e.,  j/ (a  + b)  ( a — b). 
Ex.  3.  Construct  x=l/3  ab,  i.  e.,  (3  a)  b. 

Ex.  4.  Construct  x=l/ a2 — be,  i.e.,  a2  — (1 /be)1. 

Ex.  5.  Given  a line  denoted  by  1.  construct  -j/3;  also  }\/5. 
Ex.  6.  Divide  a line  into  three  parts  proportional  to  2,  ■£,  f. 
Ex.  7,  Divide  a line  harmonically  in  the  ratio  3:6. 


EXERCISES.  PROBLEMS 


227 


Ex.  8.  Divide  one  side  of  a triangle  into  segments  proportional  to 
the  other  two  sides. 

Ex.  9.  Divide  aline  into  segments  in  the  ratio  1 : i/2. 

Ex.  10.  Given  a point  P in  the  side  AB  of  a triangle  ABC,  draw 
a line  from  P to  AC  produced  so  that  the  line  drawn  may  be  bisected 
by  BC. 

[Sug.  Suppose  the  required  line,  PQB,  drawn  meeting  BC  in  Q 
and  AC  in  B.  From  P draw  PL  [|  AC.  Compare  the  APLQ  and  QBC.  ] 

Ex.  11.  Through  a given  point  P in  the  arc  ^ ^ 

subtended  by  the  chord  AB  draw  a chord  which  / N. 

shall  be  bisected  by  AB.  I O \p 

[Sug.  Suppose  the  required  chord  drawn,  viz.,  \ ;\  /J 

PQB.  Join  the  center  0 with  P and  Q.  What  kind  / Q / ^ 

of  an  angle  is  OQP,  etc.?]  

Ex.  12.  In  an  obtuse  triangle  draw  a line  from  the  vertex  of  the 
obtuse  angle  to  the  opposite  side  which  shall  be  a mean  proportional 
between  the  segments  of  the  opposite  side. 

[Sug.  Circumscribe  a circle  about  the  triangle  and  reduce  the 
problem  to  the  preceding  Ex.] 

Ex.  13.  Find  a point  P in  the  arc  subtended  by  the  chord  AB 
such  that  chord  PA  : chord  PB— 2 : 3. 

[Sug.  Suppose  the  required  construction  made,  and  also  the  chord 
AB  divided  in  the  ratio  2 : 3 at  the  point  Q.  How  do  the  angles  APQ 
and  QPB  compare?] 

Ex.  14.  Given  the  perimeter,  construct  a triangle  similar  to  a 
given  triangle. 

Ex.  15.  Given  the  altitude  of  a triangle,  construct  a triangle 
similar  to  a given  triangle. 

Ex.  16.  In  a given  circle  inscribe  a triangle  similar  to  a given 
triangle. 

Ex.  17.  About  a given  circle  circumscribe  a triangle  similar  to  a 
given  triangle. 

Ex.  18.  By  drawing  a line  parallel  to  one  of  the  sides  of  a given 
rectangle,  divide  the  rectangle  into  two  similar  rectangles. 


228 


BOOK  III.  PLANE  GEOMETRY 


Ex.  19.  Inscribe  a square  in  a given 
triangle. 

[Sug.  If  ABC  is  the  given  triangle,  sup- 
pose DGFE  the  required  inscribed  square. 
Join  BE  and  produce  it  tc  meet  AH  ||  BC. 
Prove  AH=AK,  etc.] 


B GK  F 


374.  The  method  of  similars  in  solving  geometrical 
problems  is  best  shown  by  the  aid  of  an  example. 

Ex.  In  the  side  BC  of  a triangle  ABC  find  a point  D such  that 
the  perpendiculars  from  it  to  the  other 
sides  shall  be  in  the  ratio  3:1. 

Construction.  At  any  point  P in  AC 
erect  a X PQ  of  any  convenient  length. 

In  a direction  X AB  draw  BQ=  i QP- 
Join  BP.  From  S draw  ST  |j  BQ.  Produce 
AT  to  D.  Then  D is  the  required  point. 

Let  the  pupil  supply  the  proof. 


EXERCISES.  CROUP  37 

PROBLEMS  SOLVED  BY  METHOD  OF  SIMILARS 

Ex.  1.  In  one  side  of  a triangle  find  a point  such  that  the  perpen- 
diculars from  it  to  the  other  two  sides  shall  be  in  the  ratio  m : n. 

Ex.  2.  Find  a point  the  perpendiculars  from  which  to  the  three 
sides  of  a given  triangle  shall  be 
[Sug.  Use  Ex.  1 twice.] 

Ex.  3.  Construct  a circle 
which  shall  touch  two  given 
lines  and  pass  through  a 
given  point. 

[Sug.  Let  OA  and  OB  be 
the  given  lines  and  P the 
given  point.  Draw  any  OB  touching  the  two  lines  (OB  at  T)  and 
intersecting  OP  produced  at  X.  Draw  the  chord  IF,  etc.] 

Ex.  4 Inscribe  a square  in  a given  semi-circle. 

[Sug.  Circumscribe  a semi-circle  about  any  given  square,  by  taking 
the  midpoint  of  the  base  of  the  square  as  a center,  and  the  line  from 
this  midpoint  to  a non-adjaeent  vertex  as  a radius,  etc.] 

Ex.  5.  Solve  Ex,  ID,  p:  228,  by  the  method  of  similars. 


EXERCISES.  PROBLEMS 


229 


375.  Algebraic  analysis  of  problems.  The  conditions  of 
a problem  may  often  be  stated  as  an  algebraic  equation;  by 
solving  the  equation , the  length  of  a desired  line  in  terms  of 
Jcnown  lines  may  then  be  obtained , and  the  problem  solved 
by  constructing  the  algebraic  expression  thus  obtained. 


i Ex.  Find  a point  P in 
i the  line  AB  such  that  AP2'— 
3 'BP2. 


Analysis  and  Construction.  Denote  AB  by  a,  AP  by  x, 
and  PR  by  a — x.  Thena;2  = 3(a — x )2. 

.'.  2j5  — 6ax=  -3a2,  and  x = ~ [[ 1 

Construct  a i/3,  whence  construct  — — ^ ; lay  off  the  line  ob- 
tained, as  AP,  on  AB\  this  gives  the  point  P of  internal  division. 

Similarly,  the  construction  of  gives  Pr,  the  point  of  external 

A 

division. 


EXERCISES.  CROUP  38 


PROBLEMS  SOLVED  BY  ALGEBRAIC  ANALYSIS 


Ex.  1.  Find  a point  P in  a given  line  MR  such  thatMP2— 2RP2. 

Ex.  2.  Construct  a right  triangle,  given  one  leg 
a and  the  projection,  b,  of  the  other  leg  on  the 
-hypotenuse. 

[Sug.  Denote  the  projection  of  a on  the  hypo- 
"\tenuse  by  x.  Then  a2=x  [x-\-b),  etc.] 

Ex.  3.  Inscribe  a square  in  a given  semicircle. 

Ex.  4.  From  a given  line  cut  off  a part  which  shall  be  a mean 
pioportional  between  the  remainder  of  the  line  and  another  given  line. 


Ex.  5.  Given  AC  and  CB  arcs  of  90°,  and  a a 
given  line.  Draw  the  chord  CQ  intersecting  AB 
in  P so  that  PQ=a. 

[Sug.  x2  — y2=r2.  ax={r  + y)  ( r — y ),  etc.]  A 

Ex.  0.  Given  the  greater  segment  of  a line  di- 
vided in  extreme  and  mean  ratio,  construct  the  line. 


C 


230 


BOOK  m.  PLANE  GEOMETRY 


EXERCISES.  GROUP  39 

PROBLEMS  SOLVED  BY  VARIOUS  METHODS 

Ex.  1.  Construct  two  lines,  given  their  sum  (a  line  AB)  and  their 
ratio  ( m : n). 

Ex.  2.  Construct  two  lines,  given  their  difference  and  their  ratio. 

Ex.  3.  Divide  a trapezoid  into  two  similar  trapezoids  by  drawing 
a line  parallel  to  the  bases. 

[Sug.  Conceive  the  figure  drawn,  and.  compare  the  ratio  of  the 
bases  in  the  two  trapezoids  formed.] 

Ex.  4.  Construct  a mean  proportional  between  two  given  lines  by 
use  of  Art.  358. 

Ex.  5.  Construct  a circle  which  shall  pass  through  two  given 
points  and  touch  a given  line. 

Ex.  6.  From  a given  point  draw  a secant  to  a circle  so  that  the 
external  segment  shall  equal  half  the  secant. 

[Sug.  Draw  a tangent  to  the  O and  use  the  algebraic  method.] 

Ex.  7.  From  a given  external  point  P,  draw  a secant  meeting  a 
circle  in  A and  B so  that  PA  : AB=m  : n. 

[Sug.  Draw  a tangent  to  the  circle  from  the  point  P and  denote 
its  length  by  t.  Denote  PA  by  mx  and  AB  by  nx.  Then  m (m  n)  3? 

, mt  , 

= f,  or  t : mx=mx  : ; — , etc.] 

m + n 

Ex.  8.  Through  a given  point  P draw  a straight  line  so  that  the 
parts  of  it,  included  between  that  point  and  perpendiculars  drawn  to 
the  line  from  two  other  given  points,  shall  be  in  a given  ratio. 

[Sug.  Join  the  last  two  points,  and  divide  the  line  between  them 
in  the  given  ratio.] 

Ex.  9.  Construct  a straight  line  so  that  the  perpendiculars  on  it 
from  three  given  points  shall  be  in  a given  ratio. 

[Sug.  Let  P,  Q,  B,  be  the  given  points  and  m : n : p the  given 
ratio.  Divide  PQ  in  the  ratio  m : n and  QB  in  the  ratio  n : p,  etc.] 

Ex.  10.  Upon  a given  line  as  hypotenuse  construct  a right  tri- 
angle one  leg  of  which  shall  be  a mean  proportional  between  the 
other  leg  and  the  t^ootenuse. 


Book  IV 


AREAS  OF  POLYGONS 

376.  A unit  of  surface  is  a square  whose  side  is  a unit 
of  length,  as  a square  inch,  a square  yard,  or  a square 
centimeter. 

f 

377.  The  area  of  a surface  is  the  number  of  units  of 
surface  which  the  given  surface  contains. 

It  is  important  for  the  student  to  grasp  firmly  the  fact  that  area 
means  not  mere  vague  largeness  of  surface,  but  that  it  is  a number. 
Being  a number,  it  can  be  resolved  into  factors,  it  may  be  determined 
as  a product  of  simpler  numbers,  and  handled  with  ease  and  precision 
in  -various  ways. 

378.  Equivalent  plane  figures  are  plane  figures  having 
equal  areas. 

Thus  two  triangles  may  have  equal  areas  (be  equivalent) 
and  yet  not  be  of  the  same  shape,  that  is,  not  be  equal 
i? congruent) . 

379.  Abbreviations.  Instead  of  "area  of  a rectangle,” 
for  example,  it  is  often  convenient  to  say  simply  "rectan- 
gle.” So  instead  of  "the  number  of  linear  units  in  the 
base,”  we  use  simply  "the  base.”  In  like  manner,  for 
"product  of  the  number  of  linear  units  in  the  base  by  the 
number  of  linear  units  in  the  altitude,”  a common  ab- 
breviation is  "product  of  the  base  by  the  altitude.” 

(231) 


232 


BOOK  IV.  PLANE  GEOMETRY 


COMPARISON  OP  RECTANGLES 
Proposition  I.  Theorem 

380.  If  two  'rectangles  have  the  same  altitude,  they  are 
to  each  other  as  their  bases . 


Given  the  rectangles  EFGH  and  ABCD,  having  their 
altitudes  EE  and  AB  equal. 

To  prove  FFGB  : ABCD  = EH  : AD. 

Case  I.  When  the  bases  are  commensurable . 

Proof.  Take  some  common  measure  of  EH  and  AD,  as 
AK,  and  let  it  be  contained  in  EH  n times  and  in  AD  m 
times. 

Hence  EH  : AD  = n : m.  (Why?) 

Through  the  points  of  division  of  the  bases  of  the  two 
rectangles  draw  lines  perpendicular  to  the  bases. 

These  lines  will  divide  EG  into  n,  and  AC  into  m small 
rectangles,  all  equal.  Art‘  163' 

Hence  EFGH : ABCD  = n : m.  (Why?) 

EFGH  ABCD  = EH  : AD.  (Why  ?) 


COMPARISON  OF  RECTANGLES 


233 


Case  II.  When  the  bases  are  incommensurable. 

B C F PG 


A D E QH 


Proof.  Divide  the  base  AD  into  any  number  of  equal . 
parts,  and  apply  one  of  these  parts  to  DR. 


It  will  be  contained  in  ER  a certain  number  of  times 
with  a remainder  QR,  less  than  the  unit  of  measure. 

Draw  QP  X ER,  meeting  EG  at  P. 

Then  EQ  and  AD  are  commensurable.  Constr. 


. EFPQ  EQ 
**  ABCD  AD 


Case  I. 


If  now  the  unit  of  measure  be  indefinitely  diminished, 
the  line  QR,  which  is  less  than  the  unit  of  measure,  will 
be  indefinitely  diminished. 

. •.  EQAERas  a limit ; EFPQAEFGR  asalimit.  Art.  251. 

t — r EFPQ  , . , , EEGR 

Hence  becomes  a variable  with  ^ as  lts 

■E  Q ER 

limit;  also  —rj  becomes  a variable  with  — — as  its  limit. 

A1)  Art.  253,  3 

But  the  variable  ^^2= the  variable  always. 

ABCD  AD  J „ , 

Case  * 

. „ ,.  .,  EEGR  ER 

the  limit  ■ 7 = the  limit  (Why?) 


ABCD 


Q.  E.  D. 


381.  Cor.  If  two  rectangles  have  equal  bases,  they  are 
to  each  other  as  their  altitudes. 


234 


BOOK  IY.  PLANE  GEOMETRY 


Proposition  II,  Theorem 

382.  The  areas  of  any  two  rectangles  are  to  each  other 
as  the  products  of  their  bases  by  their  altitudes. 


Given  the  rectangles  E and  E' , having  the  bases  b and 
b' , and  the  altitudes  a and  a',  respectively. 


To  prove 


E _ b X a 
E/~  V X a' 


Proof.  Construct  a rectangle,  S,  having  its  base  equal 
to  that  of  E,  and  its  altitude  equal  to  that  of  E' . 


Then 


E _a 
~ o'" 


Art.  381. 


Also 


Art.  380. 


Taking  the  product  of  the  corresponding  members  of 
the  two  equalities, 

E b X a 
E~  V X </ 

Q.  E.  D. 


Ex.  1.  Find  the  ratio  of  the  area  of  a rectangle  whose  dimensions 
are  12  X 8 in.  to  that  of  one  whose  dimensions  are  9 X 2 in. 

Ex.  2.  How  many  bricks,  each  8X5  in.,  will  it  take  to  cover  a 
pavement  60  X 9 ft-  * 


AREAS  OF  POLYGONS 


235 


AREAS  OF  POLYGONS 

Proposition  III.  Theorem 

383.  The  area  of  a rectangle  is  equal  to  the  product  of 
its  base  by  its  altitude. 


i\u\ 

1 

Given  the  rectangle  R,  with  a base  containing  b,  and  an 
altitude  containing  h units  of  linear  measure. 

To  prove  area  of  R = b X li. 

Proof.  Let  U be  a square  each  side  of  which  contains  1 
unit  of  linear  measure. 


Then  TJ  is  the  unit  of  surface. 

Art.  376. 

R_ b X h_hxh 

Art.  382. 

U 1X1 

It 

But  — is  the  area  of  R, 

Art.  377. 

( by  definition  of  area) . 

.*.  area  of  R = b X h . 

384.  Note.  By  use  of  this  theorem,  the  problem  of  finding  the 
area  of  a rectangle  is  reduced  to  the  simpler  problem  of  measuring 
the  two  linear  dimensions  of  the  rectangle  and  taking  their  product. 
(See  Art.  1.) 

Ex.  1.  Find,  in  square  feet,  the  area  of  a rectangle  8 yds.  long 
and  5 ft.  wide. 

Ex.  2.  The  area  of  a rectangle  is  60  sq.  ft.  and  its  altitude  is  5 ft. 
Find  the  base. 


236 


BOOK  IY.  PLANE  GEOMETRY 


Proposition  IV.  Theorem 


385.  The  area  of  a parallelogram  is  equal  to  the  product 
of  its  base  by  its  altitude. 

B F O 


A ID 

Given  the  / 7 ABCD  with  the  base  AT)  (denoted  by  b) 
and  the  altitude  DF  (denoted  by  h) . 

To  prove  area  of  ABCD  — b X h. 

Proof.  From  A draw  AE  ||  DF,  and  meeting  CB  pro- 
duced, at  E. 

Then  AE  _L  CE.  Art.  123. 

.-.  AEFD  is  a rectangle  with  base  b and  altitude  h.  (Why?) 
In  the  rt.  A AEB  and  DFC, 


AB  — DC. 


(Why?) 


AE=DF. 

A AEB  = A DFC. 


(Why  ?) 
(Why?) 


To  each  of  these  equals  add  the  figure  ABFD; 

Then  rectangle  AEFD  O CJ  AB CD.  Ax.  2. 

But  area  of  the  rectangle  AEFD  = b X h.  Art  ?83. 

area  CD  ABCD  = b X h.  (Why  ?) 

Q.  E.  D. 


386.  Cor.  1.  Parallelograms  which  have  equal  bases 
and  equal  altitudes  are  equivalent. 

387.  Cor.  2.  Parallelograms  which  have  equal  bases 
are  to  each  other  as  their  altitudes / 

Parallelograms  which  have  equal  altitudes  are  to  each 
other  as  their  bases. 

388.  Cor.  3.  Any  two  parallelograms  are  to  each  other 
as  the  products  of  their  bases  and  altitudes , 


AEEAS  OF  POLYGONS 


237 


Proposition  Y.  Theorem 


389.  The  area  of  a triangle  is  equal  to  one -half  the 
product  of  its  base  by  its  altitude. 


Given  the  A J.J3C  with  the  base  AC  (denoted  by  5) , and 
the  altitude  FB  (denoted  by  h) . 

To  prove  area  of  A ABC=§  b X h. 

Proof.  Draw  BI)  ||  AG , and  CD  ||  AB , forming  theZZ7 
ABDG. 

BG  is  a diagonal  of  ZZ7  ABDG. 

:.  A ABG=hCJ  ABDG. 
areaZZ J ABDG— b X h. 

:.  area  A ABC=h  b X h. 


Then 


But 


Art.  156. 
(Why?) 

Ax.  5. 

Q.  E.  D. 

390.  Cor.  1.  Triangles  which  have  equal  bases  and 
equal  altitudes  ( or  which  have  equal  bases  and  their  vertices 
in  a line  parallel  to  the  base)  are  equivalent. 

391.  Cor.  2.  Triangles  which  have  equal  bases  are  to 
each  other  as  their  altitudes ; 

Triangles  which  have  equal  altitudes  are  to  each  other  as 
their  bases. 

392.  Cor.  3.  Any  tivo  triangles  are  to  each  other  as  the 
products  of  their  bases  and  altitudes. 


Ex.  1.  Find  the  area  of  a parallelogram  whose  base  is  9 ft.  8 in. 
and  whose  altitude  is  2 ft.  3 in. 

Ex.  2.  Find  the  altitude  of  a triangle  whose  area  is  180  sq.  in. 
and  whose  base  is  1 ft.  3 in, 


238 


BOOK  IV.  PLANE  GEOMETRY 


Proposition  VI.  Theorem 

393.  If  a,  b,  c denote  the  sides  of  a triangle  opposite  the 
angles  A,  B,  G,  respectively,  and  s = £ (a  + b + c) , the  area 
of  the  triangle  — P's  (s  — a)  (s  — b)  (s  — c) 7 


Given  the  A ABC  with  the  sides  opposite  A A,  B and  C, 
denoted  by  a,  b and  c,  respectively,  i (a  + b + c) 
denoted  by  s,  and  A an  acute  angle. 

To  prove  area  A ABC  = Vs  ( s — a)  ( s — b)  ( s — c) . 
Proof.  Draw  the  altitude  BD  and  denote  BD  by  h. 


Then 


■ b2  + c2-2b  X AD. 


.\  2 bX  AD  = b2  + c2  — a2. 

&2+c2_a2 


AD= 


2b 


But  h2  = c2  — AD2=  (c  + AD)  (c  — AD) 

b2  + c2  — a2\  ( b2  + c2  — a2' 


= (c  + 


2b 


:)(- 

!)( 


2b 

2bc  — b2  — c2  + a2 


Art.  349. 
Axs.  2,  3. 

Ax.  4. 

Art.  347. 

Ax.  8. 


/ 2 be  + b2  + c2 

V 2b  ) \ 2b 

_ [(&  + c)2-a2]  [ a2  ( b c ) ^ J 

4 b2 

_ (6  + c + a)  (6  + c — a)  (a-\-  b — c)  (a  — b + c) 

4 6- 

Now  a+6  + c = 2s  .*.  a-|-  & — c = 2 s — 2 c,  etc. 

Hyp.,  Axs,  4,  3. 


AREAS  OF  POLYGONS 


239 


:.h?= 


2s (2s — 2a)  (2s — 2c)  (2s — 2b)  16s (s — a)  (s — b)  (s — c) 


4 62  4 62 

^ 2l/s  (s  — a)  (s  — 5)  (s  — c) 


But  area  A ABC—  \ b X 7i. 


Art.  390. 


area  A ABC—  Vs  ( s — a)  (s  — b)  (s  — c) . Ax.  8. 

).  E.  D. 


Proposition  VII.  Theorem 

394.  The  area  of  a trapezoid  is  equal  to  one-lialf  the 
sum  of  its  bases  multiplied  by  its  altitude. 


Given  the  trapezoid  ABCB  with  the  bases  AT)  and  BC 
(denoted  by  b and  b') , and  the  altitude  FB  (denoted  by  h) . 

To  prove  area  of  ABCB  = £ (6  + 6')  X h. 

Proof.  Draw  the  diagonal  BB. 

Then  area  of  A ABB  — \ b X h.  (Why?) 

And  area  of  A BCB  = $ b'  X h.  (Why?) 

Adding,  area  of  ABCB=%  (b'  + b)  h.  (Why?) 

Q.  E.  D. 

395.  Cor.  The  area  of  a trapezoid  equals  the  product  of 
the  median  of  the  trapezoid  by  the  altitude.  For  the  median  of 

a trapezoid  equals  one-half  the  sum  of  the  bases  (Art.  179). 


240 


BOOK  IV.  PLANE  GEOMETRY 


396.  Scholium.  The  area  of  a polygon  of  four  or  more 
sides  can  usually  be  found  in  one  of  several  ways;  as,  by 
dividing  the  polygon  into  triangles  and  taking  the  sum  of  the 
areas  of  the  triangles;  or,  by  drawing  the  longest  diagonal 
of  the  polygon  and  drawing 

perpendiculars  to  this  diago-  /|\  

nal  from  the  vertices  which  it  s'  j X — ! X 

does  not  meet , and  obtaining  <X_ , i 1 X- X 

the  sum  of  the  areas  of  the  i i . 

triangles  and  trapezoids  thus  x\  j j 

formed.  ^ ' 


Ex.  1.  Find  the  area  of  a parallelogram  whose  base  is  1 yd.,  and 
whose  altitude  is  1 ft. 

Ex.  2.  Find  the  area  of  a triangle  whose  sides  are  5,  6,  and  7 in. 

Ex.  3.  Find  the  area  of  a trapezoid  whose  bases  are  18  and  10  in., 
and  whose  altitude  is  6 in. 

Ex.  4.  If  the  area  of  a trapezoid  is  135,  and  its  bases  are  12  and 
18,  find  its  altitude. 

Ex.  5.  The  measurement  of  the  area  of  a parallelogram  reduces 
to  the  measurement  of  what  two  straight  lines  ? 

Ex.  6.  The  measurement  of  the  area  of  a triangle  reduces  to  the 
measurement  of  what  lines  ? 

Ex.  7.  The  measurement  of  the  area  of  a trapezoid  reduces  to  the 
j measurement  of  what  lines  ? 

Ex.  8.  Prove  the  theorem  of  Art.  385  by  drawing  perpendiculars 
from  £ and  C,  instead  of  from  A and  D. 


COMPARISON  OF  POLYGONS 


241 


COMPARISON  OF  POLYGONS 
Proposition  VIII.  Theorem 

397.  If  two  triangles  have  an  angle  of  one  equal  to  an 
•angle  of  the  other,  their  areas  are  to  each  other  as  the  pro- 
ducts of  the  sides  including  the  equal  angles. 


A 


Given  the  A ABC  and  ABF  having  Z1  in  common/ 

A ABC  ABXAC 

ToproTe  NYR5-=IxTxTF 

Proof.  Draw  the  line  BC. 

Then  the  A ABC  and  ABC  may  be  regarded  as  having 
their  bases  in  the  line  AB,  and  as  having  the  common 
vertex  C. 

A ABC  AB 
A ABC  AB 


Art.  391. 


In  like  manner 


A ABC 


AC 

AF 


(Why?) 


A ABF 

Multiplying  the  corresponding  members  of  these  equali- 
A ABC_  AB  X AC 

tieS’  A ABF  AB.XAF  Ax' 4’ 

0.  E.  D. 


Ex.  In  the  above  figure,  if  AB— 12,  AC=  18,  AD= 30,  and  AF—  32, 
find  the  ratio  of  the  areas  of  the  A.  ABC  and  ABF. 


P 


242 


BOOK  IV.  PLANE  GEOMETRY 


Proposition  IX.  Theorem 

398.  The  areas  of  two  similar  triangles  are  to  each  other 
as  the  squares  of  any  two  homologous  sides. 


Given  the  similar  A ABC  and  A'B'C'  with  AB  andJ/i?' 
homologous  sides. 

A ABC  AB 2 


To  prove 


A A'B'C'  AAT 
Proof.  Draw  the  homologous  altitudes  CF  and  C'F . 


Then 


A ABC  ABXCF 


AB  CF 
;X 


Art.  392. 


A A'B' C'  A'B' X CF'  A'B'  CF' 

( any  two  A are  to  each  other  as  the  products  of  their  bases  and  altitudes). 

CF  AB 


But 


C'F'  A'B' 


Substituting  for  its  equal 


Art.  336. 


As.  8. 


A ABC 


AB  AB 

' /X 


AB- 


A A'B'C'  A'B'  A'B'  a'B'2 


Q.  D. 


Ex.  1.  If  a pair  of  homologous  sides  of  two  similar  triangles  are 
4 ft.  and  5 ft.,  find  the  ratio  of  the  areas  of  the  triangles. 

Ex.  2.  Prove  Prop.  IX  by  use  of  Prop.  VIII. 


COMPARISON  OF  POLYGONS 


243 


Proposition  X.  Theorem 


399.  The  areas  of  txvo  similar  polygons  are  to  each  other 
as  the  squares  of  any  two  homologous  sides. 

B 


Given  the  similar  polygons  ABODE  and  A'B'C'D'E' , 
with  their  areas  denoted  by  S and  S',  respectively,  and  with 
AB  and  A'B'  any  pair  of  homologous  sides. 

To  prove  S : S'=Zb2  : AAB'2 . 


Proof.  Draw  the  diagonals  AC,  AD  and  A' O',  A'D'  from 
the  homologous  vertices  A and  A' . 

These  diagonals  will  divide  the  polygons  into  similar  A. 

~ “ Art.  329. 

A ABO  AB 2 

= Art.  398. 

A A'B'C'  A'B'2 

A ABO  f Jg2\  A AGP  f HA  A APE 
Ut  AA'B'C'-\jTfj?  J A A' C'D'  \j7p'2  J AA'D'E' 

(Why  ?) 

A ABO  A ACD  _ A ADE  _ 

" A A'B'C  A A' C'D'  A A'D'E'  (Why?) 
. A ABC  + A AGP  + A ADE  _ A ABO 

’ - A A'B'C'+  A A'C'D'+  A A'D'E'  A A'B'C  * ' ‘ 


. S_  = A ABC 
•'*  S'  A A'B'C'’ 

. §L  = AE.. 

" S'  A'B'2 


Ax.  6. 
Ax.  1. 


Ex.  If  a pair  of  homologous  sides  of  two  similar  polygons  are  1 
and  2 ft.,  find  the  ratio  of  the,  areas  of  the  polygong, 


244 


BOOK  IV.  PLANE  GEOMETRY 


Proposition  XI.  Theorem 

400.  In  a right  triangle , the  square  on  the  hypotenuse 
is  equivalent  to  the  sum  of  the  squares  on  the  two  legs. 

G 


Given  AD  the  square  ou  AC  the  hypotenuse  of  the  rt. 
A ABC,  and  BF  and  BK  the  squares  on  the  legs  BA 
and  BG  respectively. 

To  prove  ADoBF- f BK. 

Proof.  Through  B draw  BL  U AE,  and  meeting  ED  in 


L.  Draw  BE  and  FC. 

Then  A ABC  and  ABG  are  rt.  A . (Why  ?) 

/.  GBCis  a straight  line.  (WhyT) 

In  the  A BAE  and  FAG,  AB  = AF,  and  AE=AG. 

(Why  •) 

Also  A BAE  = A FAG,  Ax.  A. 

( for  each  = Z BAC  + a rt.  Z ) . 

.'.  A BAE=  A FAG.  (Why!) 

But  rectangle  ALo 2 A BAE. 

[for  AL  has  the  same  base,  AE,  and  the  same  altitude,  EL,  as  A BAE) . 

Also  square  BF=^2  A FAC.  (Why?) 

.*.  rectangle  A L<?  square  BF.  Ax.  1 

Iu  like  manner  rectangle  CL<z  square  BK. 

Adding,  AL  + CL,  or  ADoBF  + BK.  Ax.  2. 


401.  Cor.  The  square  on  either  leg  of  a right  triangle 
is  equivalent  to  the  square  on  the  hypotenuse  diminished  by 
the  square  on  the  other  leg. 


CONSTRUCTION  PROBLEMS 


245 


CONSTRUCTION  PROBLEMS 

Proposition  XII.  Problem 

402.  To  construct  a square  equivalent  to  the  sum  of  two 
given  squares. 


J 

s 

s' 

/ 

/ 

z 

R 

i 

A B 


Given  two  squares  S and  S’. 

To  construct  a square  equivalent  to  S + S'. 

Construction.  Construct  a right  angle  BAC.  Art.  274. 
On  one  side  of  this  angle  take  AB  equal  to  a side  of  S, 
and  on  the  other  side  take  AC  equal  to  a side  of  S'. 

Draw  BC. 

On  a line  equal  to  BC  construct  the  square  B. 

Then  B is  the  square  required. 

Proof.  B = BC“oAB2  + A C2.  Art.  400. 

.-.UoS+S'  Ax.  8. 

Q.  E.  F. 

403.  Cor.  To  construct  a square  equivalent  to  the  sum 
tof  three  or  more  given  squares.  At  C in  the  above  figure 
erect  a line  CD  A BC  (Art.  274).  and  equal  to  a side  of 
" the  third  given  square.  Draw  DB.  DB  will  be  a side  of 
a square  equivalent  to  the  sum  of  three  given  squares,  etc. 


Ex.  1.  Construct  a square  equivalent  to  the  sum  of  two  squares 
whose  sides  are  i in.  and  1 in.,  respectively. 

Ex.  2.  By  use  of  Art.  403,  taking  a given  line  as  unity,  construct 
P2;  also  -j/3-  For  example,  construct  a line  -j/2  inches  long;  also 
one  t/3  inches  long. 


246 


BOOK  IV.  PLANE  GEOMETRY 


Proposition  XIII.  Problem 

404.  To  construct  a square  equivalent  to  the  differ?**;* 
of  two  given  squares. 


B\ 

X t~K 

Given  the  squares  8 and  S'. 

To  construct  a square  equivalent  to  the  difference  of  8 
and  S'. 

Construction.  Construct  a right  angle  BAK.  Art.  274. 

On  one  side  AB  take  AB  equal  to  a side  of  the  smaller 
given  square  S'. 

From  B as  a center  with  a radius  BG,  equal  to  a side  of 
the  larger  square,  describe  an  arc  intersecting  AK  in  0 

On  a line  equal  to  AC  construct  the  square  R . 

Then  R is  the  square  required. 

Proof.  R = AC2oBC2 — AB2,  Art.  40^. 

( the  square  on  either  leg  of  a right  tri-angle  is  equivalent  to  the  square  on 
the  hypotenuse  diminished  by  the  square  on  the  other  leg). 

Hence  R^S — S'.  Ax.  s. 

Q.  E.  F. 


Ex.  Construct  a square  equivalent  to  the  difference  of  two  squares 
whose  sides  are  1 in.  and  i in.,  respectively. 


CONSTRUCTION  PROBLEMS 


247 


Proposition  XIV.  Problem 

405.  To  construct  a square  equivalent  to  a given  paral- 
lelogram . 


Given  tlie  EH  ABCD  with  base  b and  altitude  h. 

To  construct  a square  equivalent  to  ABCD. 

Construction.  On  the  line  EG  take  EF  equal  to  h and 
EG  equal  to  b. 

On  EG  as  a diameter  construct  a semicircle.  Art.  275,  Post.  3. 

At  F erect  a _L  meeting  the  semicircumference  at  K. 

Art.  274. 

On  a line  equal  to  FK  construct  the  square  S. 

Then  8 is  the  required  square. 

Proof.  S=KF \ 

But  EF2  — b X h,  Art.  343. 

[si  from  any  point  in  a circumference  to  a diameter  is  a mean  propor- 
tional between  the  segments  of  the  diameter). 

But  area  EH  ABGB  — b X h.  ' (Why?) 

.’.  &==area  EH  ABGE. 

Q.  E.  F. 

406.  Cor.  To  construct  a square  equivalent  to  a given 
triangle,  construct  the  mean  proportional  between  the  base 
and  half  the  altitude  of  the  triangle  and  construct  a square 
on  this  mean  proportional. 


248 


BOOK  IV.  PLANE  GEOMETRY 


Proposition  XV.  Problem 


407. 

polygon. 


To  construct  a triangle  equivalent  to  a given 


C 


Given  the  pol3rgon  ABODE. 

To  construct  a triangle  equivalent  to  ABODE. 

Construction.  Let  A,  B,  C be  any  three  consecutive  ver- 
tices in  the  given  polygon. 

Draw  the  diagonal  AC. 

Draw  BE  ||  AC  (Art.  279),  and  meeting  AE  produced 
at  F. 

Draw  EC. 

In  the  polygon  FCDE  take  the  three  consecutive  ver- 
tices C,  D,  E,  and  draw  the  diagonal  CE. 

Draw  DG  ||  CE,  and  meeting  AE  produced  at  G.  Draw 
CG. 

Then  A FCG  is  the  triangle  required. 

Proof.  A ABC  =c=  AAFC,  Art.  390. 

( having  the  same  base  AC,  an<}  their  vertices  in  a line  BF  ||  the  base). 

Also  A ACE  = A ACE.  Ident. 

And  A ECD  o A ECG,  Art.  390. 

( having  the  same  base  CEand  their  vertices  in  line  DG  ||  base). 

Adding,  A ABC  + A ACE  + A ECD  o A AFC  + 
A ACE  + A ECG.  Ax.  2., 

Or  polygon  ABCDE^  A FCG. 

Q E.  F. 

408.  Cor.  To  construct  a square  equivalent  to  a given 
polygon,  use  Arts.  407  and  406. 


Ex.  Construct  a triangle  equivalent  to  a given  hexagon, 


CONSTRUCTION  PROBLEMS 


?49 


Proposition  XVI.  Problem 

409.  To  construct  a rectangle  equivalent  to  a given  square, 
and  having  the  sum  of  its  base  and  altitude  equal  to  a given 
line. 


-vrj 

i / i \ 

s 

/ i \ 

i/  ! \ 

1 1 i 

B 

A E B 


Given  the  square  8 and  the  line  AB. 

To  construct  a rectangle  equivalent  to  S,  and  having  the 
sum  of  its  base  and  altitude  equal  to  AB. 

Construction.  On  AB  as  a diameter  describe  the  semi- 
circumference  ABB.  Art.  275,  Post.  3. 

At  the  point  A erect  a _L,  AC,  equal  to  a side  of  S. 

Art.  274. 

Through  C draw  a line  ||  AB  (Art.  279),  and  meeting 
the  circumference  at  B. 

Draw  BE  A AB , Art.  273. 

Construct  the  rectangle  B with  a base  equal  to  EB  and 
an  altitude  equal  to  AE. 

Then  B is  the  rectangle  required. 

Proof.  BE2  = AEXEB.  (Why?) 

But  BE=CA.  (Why?) 

.-.  Cl2  = AE  X EB.  (Why  ?) 

Or  SoB. 

Q.  E.  F. 

410.  Cor.  The  above  problem  is  equivalent  to  the 
problem:  Given  the  sum  and  product  of  tivo  lines , to  con * 
struct  the  lines, 


250 


BOOK  IV.  PLANE  GEOMETRY 


Proposition  XVII.  Problem 

411.  To  construct  a rectangle  equivalent  to  a given  square, 
and  having  the  difference  of  its  base  and  altitude  equal  to  a 
given  line. 


C 

s 

\b 

R 

* \ / 

Given  the  square  S and  the  line  AB. 

To  construct  a rectangle  equivalent  to  S,  and  having  the 
difference  of  its  base  and  altitude  equal  to  AB. 

Construction.  On  AB  as  a diameter  describe  the  circum- 
ference ADBF.  Art.  275,  Post.  3. 

At  A erect  the  A.  AC  equal  to  a side  of  S.  Art.  274. 

Draw  CP1  through  the  center  0,  and  meeting  the  circum- 
ference at  the  points  D and  F. 

Construct  a rectangle  B with  base  equal  to  CF  and  alti- 
tude equal  to  CD. 

Then  B is  the  rectangle  required. 

Proof.  CF  : CA=CA  : CD.  Art.  358. 

.-.  CA2 — CF  X CD.  (Why?) 

/.  S<*B.  (Why?)' 

Also  the  difference  of  the  base  and  altitude  of  B = 
CF—  CD  = DF=AB. 

Q.  E.  F. 

412.  Cor.  The  above  problem  is  equivalent  to  the 
problem : Given  the  difference  and  the  product  of  two  lines . 
to  construct  the  lines. 


CONSTRUCTION  PROBLEMS 


251 


Proposition  XVIII.  Problem 


413.  To  construct  a polygon  similar  to  two  given  simi 


kir  polygons , and  equivalent 


to  their  sum. 


Given  the  similar  polygons  P and  P'. 

To  construct  a polygon  similar  to  P and  P,  and  equiva- 
lent to  their  sum. 

Construction.  Take  any  two  homologous  sides,  AB  and 
A'B ' , of  P and  P'. 

Draw  MK  J.  KL  (Art.  274),  making  MK=AB,  and  KL 


=A'B'. 

Draw  ML. 

On  A"B",  equal  to  ML,  as  a side  homologous  to  AB 
construct  the  polygon  P"  similar  to  P.  Art.  372. 

Then  P"  is  the  polygon  required. 

P AB  2 P A'B' 2 
Proof.  ~ ' 1 1 " ’ a^S0  1 Art.  399. 


,2 


A"B"  P"  A"B"° 

( the  areas  of  two  similar  polygons  are  to  each  other  as  the  squares  o' 
their  homologous  sides). 

PAP'  AB2  + MB'2 


Adding, 

But 

Or 


P" 


A"B"“ 


MK2  + KL2 


AIL2, 


AB  + A'B'=A"B"2 


AB"  + A'B'2  A"B"2 


= 1. 


A"B" 


A"B" 


As.  2. 

A 

Art.  346. 
Ax.  8. 

Ax.  8. 


\ P+  P' 
P" 


= 1 (Ax. 


1.) 


P+P~  P". 


Ax.  4. 
Q.  E.  E. 


252 


BOOK  IV.  PLANE  GEOMETPiY 


Proposition  XIX.  Problem 

, 414.  To  construct  a square  which  shall  have  a given 

ratio  to  a given  square. 


. K 

■ - T'S, 


M 

/'  J 

s 

n 

%\ 

U— « j 

!C 

S' 

a 

i 

Given  the  square  S and  the  lines  m and  n. 

To  construct  a square  which  shall  be  to  S in  the  ratio 
n : m. 

Construction.  Take  AB  equal  to  a side  of  8 and  draw 
AF,  making  a convenient  angle  with  AB. 

On  AF  take  AD  equal  to  m,  and  DF  equal  to  n. 

Draw  DB.  Draw  FC  ||  DB,  meeting  AB  produced  in  G. 

Art.  279. 

On  AC,  as  a diameter,  construct  a semicircumference 

AFC.  Art.  275j  Post.  3. 

At  B erect  a J_  BK  meeting  the  semicireumference  at  K. 

Art.  274. 


Oonstruct  a square  S'  having  a 
Then  8'  is  the  square  required. 

side  equal  to  BK,  or  x. 

Proof. 

x2=a  X b. 

(Why?) 

Also 

a : b — m -.  n. 

(Why?) 

Hence 

§_=^=c^_a 
S'  x 2 ab  b 

m 

° Ass.  8,  5. 

n 

8 _mm 
8~n  ‘ 

As.  1. 

Q.  E.  F, 


CONSTRUCTION  PROBLEMS 


253 


Proposition  XX.  Problem 

416.  To  construct  a polygon  similar  to  a given  polygon, 
and  having  a given  ratio  to  it. 


Given  the  polygon  P and  the  lines  m and  n. 

To  construct  a polygon  P'  which  shall  be  similar  to  P, 
and  be  to  P in  the  ratio  n : m. 

Construction.  Construct  a square  which  shall  be  to  the 
square  on  AB  as  n : m.  Art.  414. 

Let  A'B'  be  a side  of  this  square. 

Upon  A'B'  as  a side  homologous  to  AB  construct  a 
polygon  P'  similar  to  P.  Art.  372. 

Then  P'  is  the  polygon  required. 


Proof. 


P AB2 


(Why  t) 


P'  A'B'2/ 


Bat 


AB 2 m 
A'B'2  * 


Constr.' 


Hen<?c> 


P = m 
F n 


(Why  f ) 


Q.  E.  ? 


254 


BOOK  IY.  PLANE  GEOMETRY 


Proposition  XXI.  Problem 

41 6.  To  construct  a polygon  similar  to  one  given  polygon 
mid  equivalent  to  another  given  polygon. 


Given,  the  polygons  P and  Q. 

To  construct  a polygon  similar  to  P,  and  equivalent  to  Q. 

Construction.  Construct  a square  equivalent  to  P,  and 
let  m be  one  of  its  sides.  Art.  408. 

Construct  a square  equivalent  to  Q,  and  let  n be  one  of 
its  sides.  Art.  408. 

Construct  A'B',  the  fourth  proportional  to  m,  n , and  AB. 

Art.  366. 

On  A'B',  as  a side  homologous  to  AB,  construct  a poly- 
gon P'  similar  to  P.  Art.  372. 

Then  Pf  is  the  polygon  required. 

r , P m 2 AB2  P 

Of  — = — = -=—  • CoDstr.,  Arts.  314,399. 

Q n AiBr  P< 

P^P 
" Q P' 

P'-Q. 


Art.  305. 

«J.  F. 


EXERCISES.  THEOREMS 


255 


EXERCISES.  CROUP  40 

THEOREMS  CONCERNING  AREAS 

Ex.  1.  The  diagonals  of  a parallelogram  divide  the  parallelogram 
into  four  equivalent  triangles. 

Ex.  2.  Any  straight  line  drawn  through  the  point  of  intersection 
of  the  diagonals  of  a parallelogram  divides  the  parallelogram  into  two 
equivalent  parts. 

Ex.  3.  If,  in  the  triangle  ABC,  D and  F are  the  midpoints  of  the 
sides  AB  and  AC,  respectively,  the  area  of  ABF  equals  one -fourth  the 
area  of  ABC. 

[Sug.  Use  Art.  397.] 

Ex.  4.  If  the  midpoints  of  two  adjacent  sides  of  a parallelogram 
be  joined,  the  area  of  the  triangle  so  formed  equals  one-eighth  the 
area  of  the  parallelogram. 

Ex.  5.  If,  in  the  triangle  ABC,  D and  F are  the  midpoints  of  the 
sides  AB  and  AC,  respectively,  the  triangles  ABC  and  AFB  are  equi- 
valent. 

Ex.  6.  In  a right  triangle  show,  by  obtaining  expressions  for  the 
area  of  the  figure,  that  the  product  of  the  legs  equals  the  product  of 
the  hypotenuse  by  the  altitude  upon  the  hypotenuse. 

Ex.  7.  If  two  triangles  are  equivalent,  and  the  altitude  of  one  is 
three  times  the  altitude  of  the  other,  find  the  ratio  of  their  bases. 

Ex  8.  If  two  isosceles  triangles  have  their  legs  equal,  and  half 
of  the  base  of  one  equivalent  to  the  altitude  of  the  other,  the  tri- 
angles are  equivalent. 

Ex.  9 If  two  triangles  have  an  angle  of  one  the  supplement  of 
an  angle  of  the  other,  their  areas  are  to  each  other  as  the  products  of 
the  sides  including  these  angles. 

a b 

Ex.  10.  Prove  geometrically  that  (a+h)2  = a2+&2 

+2a6. 

Ex.  11.  Similarly  prove  (a — &)2  = a2+62 — 2a6. 

Ex.  12.  Similarly  prove  (a-\-'b){a — h)=a2 — 62. 

Ex.  13.  The  line  joining  the  midpoints  of  the  parallel  sides  of  a 
“rapezoid  divides  the  trapezoid  into  two  equivalent  parts. 


256 


BOOK  IV.  PLANE  GEOMETRY 


Ex.  14.  The  lines  joining  the  midpoint  of  one  diagonal  of 
quadrilateral  to  the  vertices  not  joined  by 
the  diagonal  divide  the  quadrilateral  into 
two  equivalent  parts. 

Ex.  15.  Given  QB  and  TS  passing 
through  P,  any  point  on  the  diagonal  AC 
of  a CD , QB  ||  AD,  and  TS  ||  AB , prove 


QBTPoPBDS. 

Ex.  16.  Given  OB=OD  ; prove  AABCo 
A ADC.  Let  the  pupil  also  state  this  as  a 
theorem  in  general  language. 


EXERCISES.  CROUP  41 

USE  OF  AUXILIARY  LINES 

Ex.  1.  Given  ABCD  a CD  and  P any 
point  inside  ABDC;  prove  APAD  -j-  APBC 
^APAB+APCD. 

Ex.  2.  The  area  of  a triangle  is  equal  to 
one  half  the  product  of  its  perimeter  by  the  radius  of  the  inscribed 
circle. 

Ex.  3 If  the  extremities  of  one  leg  of  a trapezoid  be  joined  to  the 
midpoint  of  the  other  leg,  the  middle  one  of  the  three  triangles  thus 
formed  is  equivalent  to  half  the  trapezoid. 

Ex.  4.  The  area  of  a trapezoid  is  equal  to  the  product  of  one  leg  by 
the  perpendicular  on  that  leg  from  the  midpoint  of  the  other  leg. 

Ex  5.  If  the  midpoints  of  the  sides  of  a quadri- 
lateral be  joined  in  order,  the  parallelogram  thus 
formed  is  equivalent  to  one-half  the  quadrilateral. 

Ex.  6.  A quadrilateral  is  equivalent  to  a triangle 
two  of  whose  sides  are  the  diagonals  of  the  quadrilateral,  the  angle 
included  by  these  sides  being  equal  to  one  of  the  angles  formed  by 
the  intersection  of  the  diagonals. 


EXERCISES.  THEOREMS 


257 


EXERCISES.  CROUP  42 

THEOREMS  PROVED  BY  VARIOUS  METHODS 


Ex.  1.  If  through  the  midpoint  of  one  leg  of  a trapezoid  a line  be 
drawn  parallel  to  the  other  leg  to  meet  one  base  and  the  other  base 
produced,  the  parallelogram  so  formed  is  equivalent  to  the  trapezoid. 

Ex.  2.  If  the  midpoints  of  two  sides  of  a triangle  be  joined  to  any 
point  in  the  base,  the  quadrilateral  so -formed  is  equivalent  to  half  the 
triangle. 

Ex.  3.  If  P is  any  point  on  AC  t ie  diagonal  of  a parallelogram 
ABCD,  the  triangles  APB  and  APB  an  equivalent. 


Ex.  4.  If  the  side  of  an  equilateral  triangle  be  denoted  by  a,  the 
area  of  the  triangle  equals  — • 

Ex.  5.  Find  the  ratio  of  the  areas  of  two  equilateral  triangles,  if 
the  altitude  of  one  equals  the  side  of  the  other. 


Ex.  6.  If  perpendiculars  be  drawn  from  any  point  within  an 
equilateral  triangle  to  the  three  sides,  their  sum  is  equal  to  the  alti- 
tude of  the  triangle. 


Ex.  7.  If  E is  the  intersection  of  the  diagonals  AC  and  BD  of  a. 
quadrilateral,  and  the  triangle  ABE  is  equivalent  to  the  A BEC,  then 
the  lines  AB  and  CB  are  parallel. 

Ex.  8.  If,  in  the  quadrilateral  ABCB,  the  triangles  ABC  and  ABC 
are  equivalent,  the  diagonal  AC  bisects  the  diagonal  BB. 

Ex.  9.  If  two  triangles  have  two  sides  of  one  equal  to  two  sides 
of  the  other,  and  the  included  angles  supplementary,  the  triangles  are 
equivalent. 

Ex.  10.’  If,  in  the  parallelogram  ABCB, 

F is  the  midpoint  of  the  side  BC,  and  AF 
intersects  BB  in  E,  the  triangle  BKF= the 
parallelogram  ABCB. 

Ex.  11.  Given  PQ  ||  AC,  and  PE  ||  AB; 
prove  A QAR  a mean  proportional  between 

Ex.  12.  P is  any  point  in  the  side  BC 
of  the  parallelogram  ABCB  and  BP  pro- 
duced meets  AB  produced  in  Q.  Show  that 
the  triangles  BPA  and  CPQ  are  equivalent. 


258 


BOOK  IV.  PLANE  GEOMETRY 


EXERCISES.  CROUP  43 

PROBLEMS  IN  CONSTRUCTING  AREAS 

Ex.  1.  Construct  a square  having  twice  the  area  of  a given  square. 

Ex.  2.  Construct  a square  having  three  times  the  area  of  a given 
square. 

Ex.  3.  Construct  a square  equivalent  to  the  sum  of  three  given 
squares. 

Ex.  4.  Transform  a given  triangle  into  an  equivalent  isosceles 
triangle  having  the  same  base. 

Ex.  5.  Transform  a given  triangle  into  an  equivalent  triangle 
having  the  same  base,  but  having  a given  angle  adjacent  to  the  base. 

Ex.  6.  Transform  a triangle  into  an  equivalent  triangle  with  the 
same  base,  hut  naving  another  given  side. 

Ex.  7.  Transform  a parallelogram  into  an  equivalent  parallelo- 
gram having  the  same  base,  hut  containing  a given  angle. 

Ex.  8.  Construct  a triangle  similar  to  a given  triangle  and  con- 
taining twice  the  area. 

To  construct  a similar  triangle  containing  five  times  the  area ; how 
is  the  construction  changed  1 

Ex.  9.  Bisect  a given  triangle  by  a line  parallel  to  the  base. 

Ex.  10.  Construct  a polygon  similar  to  two  given  similar  polygons, 
and  equivalent  to  their  difference. 

Ex.  11.  Draw  a line  parallel  to  one  side  of  a given  rectangle,  and 
cutting  off  five-sevenths  of  the  area. 

Ex.  12.  Bisect  a parallelogram  by  a line  perpendicular  to  the  base. 

Ex.  13.  Through  any  given  point  draw  a line  bisecting  a given 
parallelogram. 

Ex.  14.  Construct  a triangle  equivalent  to 
a given  triangle,  ABC,  having  a given  base 
AD,  hut  the  /.BAC  adjacent  to  the  base  un- 
changed. 

[Sug,  Draw  CF  ||  D£;  etc.] 


F 


D G 


EXERCISES.  PROBLEMS 


259 


Ex.  15.  Transform  a parallelogram  into  an  equivalent  parallelo- 
gram having  a given  base,  but  the  angle  adjacent  to  the  base  unchanged. 

Ex.  16.  Transform  a given  triangle  into  an  eqmivalent  right 
triangle  having  a given  leg. 

[Sug.  Use  Ex.  14,  then  Ex.  5.] 

Ex.  17.  Transform  a given  triangle  into  an  equivalent  right  tri- 
angle having  a given  hypotenuse.  i 

[Sug.  Find  the  altitude  upon  the  hypotenuse  of  the  new  triangle 
by  finding  the  fourth  proportional  to  what  three  lines  f ] 

Ex.  18.  Transform  a re-entrant  pentagon  into  an 
equivalent  triangle. 

Ex.  19.  Transform  a given  triangle  into  an 
equivalent  equilateral  triangle. 

[Sug.  See  Art.  416.] 

Ex.  20.  Bisect  a triangle  by  a line  perpendicular  to  one  of  its  sides. 

[Sug.  See  Art.  416.] 

Ex.  21.  Construct  a square  equivalent  to  two -thirds  of  a given 
square. 

EXERCISES.  CROUP  44 

PROBLEMS  SOLVED  BY  ALGEBRAIC  ANALYSIS 

Ex.  1.  Transform  a given  rectangle  into  an  equivalent  rectangle 
with  a given  base. 

Ex.  2.  Transform  a given  square  into  a right  triangle  having  n 
given  leg. 

Ex.  3.  Transform  a given  triangle  into  an  equivalent  isosceles 
right  triangle. 

Ex.  4.  Draw  a line  cutting  off  from  a given  triangle  an  isosceles 
triangle  equivalent  to  one-half  the  given  triangle. 

[Sug.  Use  Art.  397.] 

Ex.  5.  Through  a given  point  in  one  side  of  a given  triangle 
draw  a line  bisecting  the  area  of  the  triangle. 

Ex.  6.  Transform  a given  square  into  a rectangle  which  shall  have 
three  times  the  perimeter  of  the  given  square. 


260 


BOOK  IV.  PLANE  GEOMETRY 


EXERCISES.  CROUP  45 

PROBLEMS  SOLVED  BY  VARIOUS  METHODS 

Ex.  1.  Bisect  a given  parallelogram  by  a line  parallel  to  the  base 

Ex.  2.  Transform  a parallelogram  into  an  equivalent  parallelo- 
gram having  the  same  base  and  a given  side  adjacent  to  the  base. 

Ex.  3.  Construct  a square  which  shall  contain  four-sevenths  of 
the  area  of  a given  square. 

Ex.  4.  In  two  different  ways  construct  a square  having  three 
times  the  area  of  a given  square. 

Ex.  5.  Trisect  a given  triangle  by  lines  parallel  to  the  base. 

Ex.  6.  Find  a point  within  a triangle  such  that  lines  drawn  from 
it  to  the  vertices  trisect  the  area. 

Ex.  7.  Find  a point  within  a triangle  such  that  lines  drawn  from 
it  to  the  three  vertices  divide  the  area  into  parts  which  shall  have 
The  ratio  2:3:4. 

[Sug.  If  one  of  the  small  A contains  f the  area  of  original  A,  a 
line  through  its  vertex  cuts  off  f the  altitude,  etc.] 

Ex.  8.  Divide  a triangle  into  three  equivalent  parts  by  lines 
through  a given  vertex. 

Ex.  9.  Divide  a triangle  into  three  equivalent  parts  by  lines  drawn 
through  a given  point  P in  one  of  the  sides. 

[Sug.  Use  Art.  397.] 

Ex.  10.  Divide  a given  quadrilateral  into  three  equivalent  parts 
by  lines  drawn  through  a given  vertex. 

Ex.  11.  Through  a given  point  in  the 
base  of  a trapezoid  draw  a line  bisecting  the 
area  of  the  trapezoid. 

Ex.  12.  Bisect  the  area  of  a trapezoid 
by  a line  drawn  parallel  to  the  bases. 

[Sug.  Construct  A GEF  similar  to  A 

ABG  and  equivalent  to  i sum  ©£  what  two 
AT] 


G 


Book  Y 

REGULAR  POLYGONS.  MEASUREMENT  OF 
THE  CIRCLE 


417.  Def.  A regular  polygon  is  a polygon  that  is  both 
equilateral  and  equiangular. 

Proposition  I.  Theorem 

418.  An  equilateral  polygon  that  is  inscribed  in  a circle 
is  also  equiangular  and  regular. 


Giveri  ABC . . . K an  inscribed  polygon,  with  its  sides 
AB,  BC,  CD,  etc.,  equal. 

To  prove  the  polygon  ABC . . . K equiangular  and 
regular. 

Proof.  Arc  AB  = ava  BC=avc  CD,  etc.  Art.  218. 

.'.  arc  A B C=  arc  BCD  = arc  CDE,  etc.  As.  2. 

.’.  Z ABC  = /.BCD  = Z CDE,  etc.,  Art.  260. 

(all  A inscribed  in  the  same  segment,  or  in  equal  segments,  are  equal). 

:.  the  polygon  ABC . . . K is  equiangular, 
the  polygon  ABC  ...  A is  regular. 


(261) 


Art.  417. 
Q.  E.  D. 


262 


BOOK  V.  PLANE  -GEOMETRY 


419.  Coe.  1.  If  the  arcs  subtended  by  the  sides  of  a 
regular  inscribed  polygon  be  bisected,  and  each  point  of 
bisection  be  joined  to  the  nearest  vertices  of  the  polygon,  a 
regular  inscribed  polygon  of  double  the  number  of  sides  is 
formed. 

420.  Cor.  2.  The  perimeter  of  an  inscribed  polygon  is 
less  than  the  perimeter  of  an  inscribed  polygon  of  double  the 
number  of  sides. 


421.  A circle  may  be  circumscribed  about,  and  a circle 
may  be  inscribed  in,  any  regular  polygon. 


Given  the  regular  polygon  ABODE. 

To  prove  that  a O may  be  circumscribed  about,  or  in- 
scribed in,  ABODE. 

Proof.  I.  Through  A,  B and  C (Fig.  1),  any  three  suc- 
cessive vertices  of  the  polygon  ABODE,  pass  a circum- 
ference. Art.  235. 

Let  0 be  the  center  of  this  circumference. 

Draw  the  radii  OA,  OB,  00.  Also  draw  the  line  OD. 

Then,  in  A OBC,  0B  = 00.  ( Why  ?) 


Proposition  II.  Theorem 


d 


n 


,cr 


Z OBC  = Z OCB. 


(Why  f ) 


KEGULAE  POLYGONS 


263 


But  /.ABC  = /.BCD,  Art.  417. 

( being  A of  a regular  polygon). 

Subtracting,  Z OBA  — / OCD.  (Why?) 

Hence,  in  the  A OAB  and  OCD, 

OB=OC.  (Why?) 

AB—  CD.  (Why?) 

Z OBA  = / OCD , 

(just  pj'oved) . 

A ABO=  AOCD.  (Why?) 

OD=OA.  (Why?) 


Hence  the  circumference  which  passes  through  the 
vertices  A,  B and  C,  will  also  pass  through  the  vertex  D. 

In  like  manner,  it  may  be  proved  that  this  circumference 
will  pass  through  the  vertex  E. 

Hence  a circle  described  with  0 as  a center,  and  OA  as 
a radius,  will  be  circumscribed  about  the  given  polygon. 

II,  The  sides  of  the  polygon  ABCDE  (Fig.  2)  are  equal 
chords  in  the  circle  0. 

Hence  they  are  equidistant  from  the  center.  Art.  226. 

.'.  a circle  described  with  0 as  a center,  and  the  distance 
from  0 to  one  of  the  sides  of  the  polygon  as  a radius,  will 
be  inscribed  in  the  given  polygon. 

Q.  E.  D. 

422.  Def.  The  center  of  a regular  polygon  is  the  com- 
mon center  of  the  inscribed  and  circumscribed  circles,  as 
the  point  0 in  the  above  figure. 

423.  Def.  The  radius  of  a regular  polygon  is  the  radius 
of  the  circumscribed  circle,  as  OA  in  the  above  figure. 

424.  Def.  The  apothem  of  a regular  polygon  is  the 
radius  of  the  inscribed  circle. 


264 


BOOK  Y.  PLANE  GEOMETRY 


425.  Def.  The  angle  at  the  center  of  a regular  polygon 

is  the  angle  between  two  radii  drawn  to  the  extremities  of 
any  side,  as  the  angle  AOB. 

426.  Cor.  The  angle  at  the  center  of  a regular  polygon 
is  equal  to  four  right  angles  divided  by  the  number  of  sides. 

Hence,  if  n denote  the  number  of  sides  in  the  polj'gon, 

, the  angle  at  the  center  of  a regular  polygon  equals  ^ r^'  ^ ; 
l n 

also  the  angle  between  an  apothem  and  the  nearest  radius, 

in  a regular  polygon  of  n sides,  equals  1 ^ ^ • 


Proposition  III.  Theorem 

427.  If  the  circumference  of  a circle  be  divided  into  any 
number  of  equal  arcs, 

I.  The  chords  of  these  arcs  form  a regular  inscribed 
polygon; 

II.  Tangents  to  the  circumference  at  the  points  of  division 
form  a regular  circumscribed  polygon. 


Given  the  circumference  ABC,  divided  into  the  equal 
arcs  AB,  BC,  CD,  etc.,  the  chords  AB,  BC,  etc.,  and  PQ, 
QB,  etc.,  lines  tangent  to  the  circle  at  B,  C,  etc. 

To  prove  ABODE  a regular  inscribed  polygon,  and 
PQBST  a regular  circumscribed  polygon. 


REGULAR  POLYGONS 


265 


Proof.  I.  The  chords  AB,  BC,  CD,  etc.,  are  equal. 

(Why  ?) 

polygon  ABODE  is  equiangular  and  regular.  Art.  418. 


II.  In  the  A APB,  BQG,  CRD,  etc., 

AB  — BC=  CD,  etc.  (Why  f) 

Also  Z PAB  = Z PBA  = Z QBC  = Z QCB  = Z RCD,  etc. 

Art.  264. 

(each  beinq  measured  by  half  of  one  of  the  equal  arcs 
AB,  BC,  CD,  etc.)  - ' 

;.  A APB,  BQC,  CRD,  etc.,  are  equal,  isosceles  triangles. 


(Why?) 

.*.  ZP  = ZQ  = ZR,  etc.  (Why?) 

And  AP—PB=BQ—  QC,  etc.  (Why?) 

PQ—QR  = RS,  etc.  Ax.  4. 

.*.  PQRST  is  a regular  polygon.  Art.  417. 


Q.  E.  D. 

428.  Cor.  1.  If  the  arcs  AB,  BC,  CD,  etc.,  he  bisected, 
and  a tangent  he  drawn  at  each  point  of  bisection,  a circum- 
scribed regular  polygon  of  double  the  number  of  sides  of 
PQRST  will  be  formed. 

429.  Cor.  2.  The  perimeter  of  a circumscribed  regular 
polygon  is  greater  than  that  of  a circumscribed  regular  poly- 
gon of  double  the  number  of  sides. 


Ex.  1.  Find  the  number  of  degrees  in  the  central  angle  of  a 
regular  pentagon.  Of  a regular  hexagon.  Of  a square. 

Ex.  2.  What  is  the  short  name  for  an  inscribed  equilateral  quad- 
rilateral ? 


BOOK  V.  PLANE  GEOMETRY 


266 

Proposition  IV.  Theorem 

430.  Tangents  to  a circle  at  the  midpoints  of  the  arcs 
subtended  by  the  sides  of  a regular  inscribed  polygon  form  a 
regular  circumscribed  polygon  whose  sides  are  parallel  to 
the  corresponding  sides  of  the  inscribed  polygon. 


A 


Given  the  regular  polygon  ABODE  inscribed  in  the  O 
ACD-,  P,  Q,  B,  etc.,  the  midpoints  of  the  arcs  AB,  BC, 
CD,  etc.;  and  A'B',  B'C' , CD’,  etc.,  tangents  to  the  circle 
at  P,  Q,  R,  etc. 

To  prove  A'B'C'D'E'  a regular  polygon  with  its  sides  1 1 
corresponding  sides  of  the  polygon  ABODE. 

Proof.  The  arcs  AB,  BO,  CD,  etc.,  are  equal.  Art.  218. 
.'.  the  arcs  AP,  PB,  BQ,  QC,  etc.,  are  equal.  Ax.  5. 


.'.  the  arcs  PQ,  QB,  RS,  etc.,  are  equal.  Ax.  4. 
A'B'C'D'E'  is  a regular  polygon,  Art.  427. 
(if  the  circumference  of  a O he  divided,  etc.). 

Side  AB  J_  OP.  Art.  113. 

A'B1  _L  OP.  (Why?) 

.*.  AB  ||  A'B'.  (Why?) 


In  like  manner,  each  pair  of  homologous  sides  in  the 
two  polygons  is  parallel. 

Q.  E.  D. 

431.  Cor.  Homologous  radii  of  an  inscribed  and  a cir- 
cumscribed regular  polygon,  whose  sides  are  parallel,  coin- 
cide in  direction.  Thus,  in  the  above  figure  A POA  and 

POA'  each  ="■■  I't-~ 


n 


OA  and  OA'  coincide  in  direction. 

Art.  426. 


BE GULAK  POLYGONS 


267 


Proposition  V.  Theorem 

432.  Two  regular  polygons  of  the  same  number  of  sides 
are  similar. 


Given  K and  K'  two  regular  polygons,  each  of  n sides. 

To  prove  K and  K ' similar. 

_ , „ . . „ „ (n  — 2)2  rt.  A 

Proof.  Each  Z of  K = > Art.  174. 

n 

/■  -7  7 , , , (n  — 2)  2 rt.  A . 

(m  an  equiangular  polygon  of  n sides , each  Z = 


Similarly  each  Z of  K!~ 


(n  — 2)  2 rt,  Z 


Hence  K and  K'  are  mutually  equiangular.  Ax,  I,  Art.  igs. 

AB  . 

— 1.  Art.  417,  Ax.  5. 


Also 

And 


AB  — BC 
A'B'  — B'G' 

. = ^ or 


BC 
A'B' 


= 1. 


B'G' 

AB  BG 


BG  B'G'  A'B'  B'G' 

, ...  BG  CD  DE 

In  like  manner  T.,  etc. 


(Why?) 

(Why?) 


B'G'  C'D'  D'Er 
Hence  K and  K'  have  their  homologous  sides  propor- 
ional. 

Hence  E and  K'  are  similar.  Art.  321. 

Q.  E.  D. 


433.  Cor.  The  areas  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  squares  of  any 
two  homologous  sides, 


268 


BOOK  V.  PLANE  GEOMETRY 


Proposition  VI.  Theorem 


434.  I.  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  radii  of  their 
circumscribed  circles,  or  as  the  radii  of  their  inscribed 


circles ; 

II.  Their  areas  are  to  each 
radii. 


D 


other  as  the  squares  of  these 


Given  AC  and  A' C'  two  regular  polygons,  each  of  n 
sides,  with  centers  0 and  O',  and  with  perimeters  denoted 
bjr  P and  P',  radii  by  R and  R',  and  apothems  by  r and  r', 
respectively. 

To  prove.  I.  P : P'  = R : R'  = r : r'. 

II.  Area  AC  : area  A'C'  = R2  : Rr2=r2  : r n. 

Proof.  I.  The  polygons  AC  and  A'C'  are  similar.  Art.  432. 


Hence  P : P'  = AB  : A'B' . Art.  341. 

But,  in  the  A OAB  and  O' A'B', 

Z AOB  = Z A'O'B',  [for  each  Z ) . Art.  426. 

Also  OA  : OB  = O' A'  : O'B',  (for  each  A is  isosceles). 

A OAB  and  O' A'B'  are  similar.  Art.  327. 

A AB  : A'B'=  OA  : O’ A'.  Art.  321. 

And  AB  : A'B'  — OL  : O'B'.  Art.  338. 

/.  P : P'=OA  : O'  A'  — OL  : O’U . Ax.  1. 

Or  P : P'=R  : R'  = r : r'. 

II.  Area  AC  : area  A'C'  = AB2  : A'B'2.  Art.  399. 

But  AB2  :A7B,2  = R2  : R,2  = r2  : r'2.  Art.  314. 

.*.  area  AC  ; area  A'C'  = R 2 : R'2  ; r 2 : r'2.  Ax.  1. 


Q.  £.  D. 


REGULAR  POLYGONS 


269 


Proposition  VII.  Theorem 

435.  If  the  number  of  sides  of  a regular  inscribed  poly- 
gon be  indefinitely  increased , the  apothem  of  the  polygon 
approaches  the  radius  as  a limit. 


Given  the  regular  inscribed  polygon  AB  ...  2)  of  w sides, 
with  radius  OA  and  apothem  OL. 

To  prove  that,  as  n is  indefinitely  increased,  OL  ap- 
proaches OA  as  a limit. 

Proof.  In  the  A OAL,  AL  > OA  — OL,  Art.  93. 

[any  side  of  a A is  greater  than  the  difference  between  the 
oilier  two  sides.) 

But,  as  n = 30 , 1£aO  .'.  AL  = 0.  Art.  253,  3. 

Hence  OA— OL^  0. 

0L=  OA,  or  r = B. 

Or  the  limit  of  thp  apothem  OL  is  the  radius  OA. 

Q.  E.  D. 

436.  Cor.  As  n = <x>,B2 — r2  = 0. 

For.  B2 — r2=(B-\-r ) (B — r) . But,  asn=^, 

B + r==B  + B or  2 B,  and. 22  — r = 0.  Ax.  8. 

B1  — r2  = 2B  X 0 or  0.  Ax.  8. 


Ex.  A pair  of  homologous  sides  of  two  regular  pentagons  are  2 
and  3 ft,  Find  the  ratio  of  the  areas  of  tho  polygons. 


270 


BOOK  Y.  PLANE  GEOMETRY 


Proposition  VIII.  Theorem 

437.  The  length  of  any  line  inclosing  the  circumference 
of  a circle,  and  not  passing  within  the  circumference,  is  greater 
than  the  length  of  the  circumference. 


Given  the  circumference  AKEF,  and  ABCDEF  any  line 
which  does  not  pass  within  AKEF. 

To  prove  circumference  AKEF  < perimeter  ABCDEF. 

Proof.  Let  K be  any  point  on  the  circumference  AKEF 
not  touched  by  the  line  ACDEF. 

At  K draw  a tangent  to  the  given  circle,  meeting  ACEF 
at  B and  D. 

Then  the  straight  line  BKD  < line  BCD.  Art.  15 

To  each  of  these  unequals  add  the  line  DEFAB. 

Then  line  ABKDEF  < line  ACDEF.  Ax.  5 

Hence  every  line  enveloping  the  circular  area  AKEF 
except  the  circumference  AKEF,  may  be  shortened. 

Hence  the  circumference  AKEF  is  shorter  than  any  line 
enveloping  it. 

Q.  E.  D. 


KEGULAK  POLYGONS 


271 


438.  Cor.  1.  The  circumference  of  a circle  is  less  than 
the  perimeter  of  any  polygon  circumscribed  about  the  circle. 

439.  Cor.  2.  The  circumference  of  a circle  is  greater 
'than  the  perimeter  of  any  polygon  inscribed  in  the  circle, 
for  each  side  of  an  inscribed  polygon  is  less  than  the  arc 
subtended  by  it. 

440.  Cor.  3.  The  difference  between  the  perimeters  of 
an  inscribed  and  a circumscribed  polygon  is  greater  than  the 
difference  between  either  perimeter  and  the  circumference  of 
the  circle. 


Proposition  IX.  Theorem 

441.  If  the  number  of  sides  of  a regular  inscribed,  or 
of  a regular  circumscribed  polygon  be  indefinitely  increased, 

I.  The  perimeter  of  each  polygon  approaches  the  circum- 
ference as  a limit; 

II.  The  area  of  each  polygon  approaches  the  area  of  the 
circle  as  a limit. 


Jiven  a circle  of  circumference  C and  area  A,  with  regu- 
lar inscribed  and  circumscribed  polygons,  each  of  n sides, 
with  their  perimeters  denoted  by  P and  P' , and  their  areas 
by  K and  K' , respectively. 

To  prove  that  as  n is  indefinitely  increased,  P and  P' 


272 


BOOK  V.  PLANE  GEOMETKY 


each  approaches  C as  a limit,  and  K and  K'  each  ap- 
proaches A as  a limit. 


Proof  I.  Denote  the  apothems  of  the  two  polygons  by 
R and  r. 


Then 

Hence 
Let  n = oo  ; 


P = R 
P r 

P’—P  R — r 
P r 

then  R — r = 0. 


Art.  434. 

Art.  310. 
Art.  435. 
Art.  253,  4> 


( for  the  denominator,  r,  increases  as  n increases). 

Hence  1—~>  =0  (Ax.  8).  .-.  P'  - P=  0,  (Art.  253.  4). 

But  P>—  G < P'—P  (Art.  440)  /.  P'— <7=0,  or  P = C. 
Also  C—  P < P'—P  { Art.  440)  /.  O—  PM),  or  P=C. 


TT  K'  P2  ,A  . . K' 

II.  — = — (Art.  434)  .. 

K >- 


K R2  — r2 


K 


Let  n = co  ; then  Rz  — r2  A 0. 


Art.  310. 
Art.  436. 


Art.  253,  4. 


(for  the  denominator , r-,  increases). 

T£!  

Hence  = = 0 (Ax  8).  .'.  K'  — K = 0,  (Art.  253  4). 

A 

But  K'—A  < K'—K  (Ax.  7)  /.  K' — A = 0,  or  K'  = A. 
Also  A — K < K' — K (Ax.  7)  .-.  A — AT=0,  or  K=A. 

Q.  E.  D. 


Ex.  Find  the  perimeter  and  area  of  a square  field,  one  of  whose 
sides  is  10  rods.  Find  the  same  in  a square  field,  one  of  whose  sides 
is  20  rods.  Is  it  more  economical,  therefore,  to  fence  land  in  large 
or  small  fields  ? v 


REGULAR  POLYGONS 


273 


Proposition  X.  Theorem 


442.  Two  circumferences  have  the  same  ratio  as  their 
radii,  or  as  their  diameters. 


Given  the  circles  0 and  O',  with  circumferences  denoted 
by  0 and  O',  radii  by  R and  R',  diameters  by  D and  U , re- 
spectively. 

To  prove  C : C = R : R'  = D : D'. 

Proof.  In  the  given  © let  regular  polygons  of  the  same 
number  of  sides  be  inscribed.  Denote  the  perimeters  of  the 
inscribed  polygons  by  P and  P' , respectively. 

mi  PR 

Then  — = — Art.  434. 


Hence,  by  alternation, 


R 


Pf% 

Rf' 


Art.  307. 


If,  now,  the  number  of  sides  of  the  similar  inscribed  poly- 
gons be  increased  indefinitely,  P becomes  a variable  ap- 
proaching C as  a limit.  Art.  441. 


Hence—  becomes  a variable  approaching  — as  a limit. 

R R Art.  253,  3, 

pr  C' 

Similarly,  the  variable  — approaches  — as  a limit. 

K R 

P P' 

But  the  variable  — = the  variable  — always. 

K K ' 


Also 


R_ 

R' 


2 R 
2 Rf 


C=Cf' 

R R 

C : C'  = R : Rf. 

(Art.  315.)  A — = ^-  = ~ ■ 

O 2 R'  I)' 


R 


Art.  254. 
Art.  307. 

Ax.  8. 

Q.  E.  D. 


274 


BOOK  Y.  PLANE  GEOMETRY 


Proposition  XI.  Theorem 


443.  In  every  circle,  the  ratio  of  the  circumference  to  the 
diameter  (that  is,  the  number  of  times  the  diameter  is 
1 contained  in  the  circumference)  is  the  same,  and  may  be 


denoted  by  an  appropriate  symbol  (71) . 


To  prove 
Proof. 

By  alternation 


Art.  442. 
Art.  307. 


Given  two  © with  circumferences  denoted  by  G and  C\ 
and  diameters  by  D and  D' , respectively. 

C _ C'_ 

JD  V 71 ' 

(f  = D' 

G’  D' 

C = 

D B'‘ 

0 

Hence,  in  any  given  circle,  — has  a value  equal  to  that 

G ^ 

of  — in  some  standard  circle. 

Q 

the  value  of  — is  the  same  in  all  circles. 

J 0 

.Denoting  this  constant  by  71,  in  every  circle  — —7t. 

Q.  E.  D. 

444.  Formula  for  the  circumference  iq  terms  of  the 
,-adius.  r 

We  have  — =7t  .'.  C=7tD  = 7l(2  R) . 

(7=2  ttE. 


Ex.  Find  the  circumference  of  a circle  whose  radius  is  10  inches* 


REGULAR  POLYGONS 


275 


445.  Formula  for  length  of  arc  of  a circle. 


arc 


circumference 


central  Z 
360° 


(Art.  255.)  arc  = 


central  Z 
360° 


X2nB. 


central  Z ..  _ 

arc- — — X 71  JR. 


Proposition  XII.  Theorem 

446.  The  area  of  a regular  polygon  is  equal  to  one-halj 
the  product  of  its  perimeter  by  its  apothem. 


Given  the  regular  polygon  ABODE  with  area  denoted 
oy  K,  'perimeter  by  P,  and  apothem  by  r. 

To  prove  K=§  P X r. 

Proof.  Draw  the  radii  OA,  OB,  OG,  etc.,  dividing  the 
polygon  into  as  many  A as  the  polygon  has  sides. 

All  the  A thus  formed  have  the  same  altitude,  r.  Art.20* 

/.  the  area  of  each  A=J  product  of  its  base  by  r. 

Art.  389. 

Hence  the  sum  of  the  areas  of  the  product  of  the 

sum  of  the  bases  of  the  A by  r,  or  = J P X r. 

But  the  sum  of  the  areas  of  the  A equals  the  area  of 
the  polygon.  Ax.  6. 

Hence  K=hPXr.  Ax.  8. 

Q.  E.  D. 

447.  Def.  Similar  sectors  are  sectors  in  different  cir- 
cles which  have  equal  angles  at  the  center. 

448.  Def.  Similar  segments  are  segments  in  different 
circles  whose  arcs  subtend  equal  angles  at  the  center. 


276 


BOOK  V.  PLANE  GEOMETRY 


Proposition  XIII.  Theorem 

449.  The  area  of  a circle  is  equal  to  one-half  the  -prod 
uct  of  its  circumference  hy  its  radius. 


Given  a O with  circumference  denoted  by  C,  radius  by 
R,  and  area  by  K. 

To  prove  K=  £ CX  R. 

Proof.  Circumscribe  a regular  polygon  about  the  given 
circle,  and  denote  its  perimeter  by  P and  its  area  by  K'. 

In  this  case  the  apothem  of  the  regular  polygon  is  R. 
Hence  K'  = % P X R.  Art.  446. 

Let  the  number  of  sides  of  the  circumscribed  polygon 
be  inci’eased  indefinitely;  then 

K!  becomes  a variable  approaching  K as  its  limit ; Art.  441. 
P becomes  a variable  approaching  C as  its  limit;  Art.  441. 
And  variable  $PxR  approaches  £ Cx  R as  a limit.  Art.  253. 2 
But  K'  — \ P X R always.  Art  44f> 

Hence  K — \ C X R.  Art  254 

Q.  E.  D. 

450.  Formula  for  the  area  of  a circle  in  terms  of  the 
adius  R. 

K=$  CX  R;  but  C—2nR.  Art.  444. 

.*.  =-H2  7tP)P.  Or  K—  7lR~.  Ax.  8. 

nP>~ 

Again  R=i  B K———-  Ax.  8. 

4 

451.  Cor.  1.  The  area  of  a circle  is  equal  to  the  square 
of  the  radius,  multiplied  by  n;  or  to  one- fourth  the  square  of 
the  diameter,  multiplied  by  n. 


MEASUREMENT  OF  THE  CIRCLE 


277 


452.  Cor.  2.  The  areas  of  two  circles  are  to  each  other 
as  the  squares  of  their  radii,  or  as  the  squares  of  their  diam- 
eters. 


For 


E 7tB 2 R2  i K InD2  D2 
‘E~nE2~Ri2'  also  K,= lnD,2=lF2 


453.  The  area  of  a sector  is  equal  to  one-lialf  the  prod- 
uct of  its  radius  by  its  arc. 

( central  Z 


Hence  area  of  sector  = J ^ — X nR  ^ R. 

Or  area  of  sector  = — -^7^7— X 7tR2. 

ob(J 


Art.  445. 


454.  Cor.  3.  Similar  sectors  are  to  each  other  as  the 
squares  of  their  radii. 


Ex.  1.  The  measurement  of  the  area  of  a circle  can  be  reduced 
to  the  measurement  of  the  length  of  what  single  straight  line  ? Can 
it  be  reduced  to  the  measurement  of  any  other  single  straight  line  1 
to  the  measurement  of  a single  curved  line  ? 


Find  the  area  of  a circle, 

Ex.  2.  “Whose  radius  is  10  ft. 

Ex.  3.  “Whose  diameter  is  10  ft 
Ex.  4.  “Whose  radius  is  6 ft. 
Ex.  5.  Whose  radius  is  46  ft. 


Ex.  6.  Whose  radius  is  26  ft. 

Ex.  7.  Whose  radius  is  B;  iB; 
2 B. 

Ex.  8.  Whose  radius  is  Bf3. 
Ex.  9.  Whose  diameter  is  iEi/ 2 


Ex.  10.  If  the  radius  of  one  circle  is  10  times  as  great  as  the  ra- 
dius of  another  circle,  how  do  their  areas  compare  ? Also  how  do 
their  circumferences  compare  f 


Ex.  11.  A wheel  with  6 cogs  is  geared  to  a wheel  with  48  cogs. 
How  many  revolutions  will  the  smaller  wheel  make  while  the  larger 
wheel  revolves  once  ? 


Ex.  12.  A 2 in.  pipe  will  discharge  how  much  more  water  ia  a 
given  time  than  a 1 in.  pipe  ? 


278 


BOOK  Y.  PLANE  GEOMETRY 


Proposition  XIV.  Theorem 


455.  The  areas  of  two  similar  segments  are  to  each  other 
as  the  squares  of  their  radii. 

O 


Given  the  similar  segments  ABC  and  A'B'C'  in  circles 
whose  radii  are  B and  B'. 


To  prove  segment  ABC  : segment  A'B'C'  = B 2 : B'2. 
Proof.  Draw  the  radii  OA,  OB , O'A',  O’B' . 

Then  the  sectors  OAB  and  O'A'B'  are  similar. 


Arts.  448,  447. 

And  A OAB  and  O'A'B'  are  similar.  Art.  327. 


. sector  OAB  _B2 _ A OAB 
" sector  0'A'B'~  B'2'  and  A O'A'B' 

. sector  OAB  A OAB 
“ sector  O'A'B'-  A O'A'B'' 


B?_ 

B'2 

Art.  454,  398. 
(Why  ?) 


. sector  OAB  sector  O'A'B' 
A OAB  ~ A O'A'B' 


(Why  ft) 


. sector  OAB  — A OAB  sector  O'A'B' — A O'A'B' 
A OAB  ~ A O'A'B' 


Or 


segment  ABC _ segment  A'B'C' 
A OAB  ~ A O'A'B' 

segment  ABC  / A OAB  \ _B2 
segment  A'B'C'  VA  O'A'B')  B'2 


(Why?) 


Arts.  307,  398 


Q.  E.  D. 


CONSTRUCTION  PROBLEMS 


279 


CONSTRUCTION  PROBLEMS 

Proposition  XV.  Problem 

456.  To  inscribe  a square  in  a given  circle. 

C 


Let  the  pupil  supply  a solution. 

457.  Cor.  Bij  bisecting  the  arcs  AC,  CB,  BD,  etc., 
and  drawing  chords,  a regular  octagon  may  be  inscribed  in 
the  circle;  by  repeating  the  process,  regular  polygons  of  16, 
32,  64,  . . . and  2n  sides  may  be  inscribed,  where  n is  a 
positive  integer  greater  than  1. 

How  can  a regular  polygon  of  2n  sides  be  circumscribed 
about  a given  circle  ? 


Proposition  XVI.  Problem 

458.  To  inscribe  a regular  hexagon  in  a given  circle. 


Let  the  pupil  supply  a solution. 


280 


BOOK  V.  PLANE  GEOMETEY 


459.  Cor.  1.  The  side  of  a regular  inscribed  hexagon 
equals  the  radius  of  the  circle. 

460.  Cor.  2.  By  joining  the  alternate  vertices  of  a 
regular  inscribed  hexagon , an  equilateral  triangle  can  be  in- 
scribed in  a given  circle. 

461.  Cor.  3.  By  bisecting  the  arcs  AB,  BC,  CD,  etc., 

! and  draiving  chords,  a regular  polygon  of  12  sides  can  be 

inscribed  in  a given  circle;  by  repeating  the  process,  regular 
polygons  of  24,  48,  . . . 3 X 2*  sides  can  be  inscribed. 

Similarly,  how  can  a regular  polygon  of  3X2”  sides 
be  circumscribed  about  a given  circle  ? 


Proposition  XVII.  Problem 

462.  To  inscribe  a regular  decagon  in  a given  circle. 


Given  the  circle  0. 

To  inscribe  a regular  decagon  in  the  given  O. 

Construction.  Draw  any  radius  OA,  and  divide  OA  in 
extreme  and  mean  ratio  at  E,  OK  being  the  greater  seg- 
ment. Art.  371. 

With  A as  a center  and  OK  as  a radius,  describe  an  arc 
cutting  the  given  circumference  at  B. 

Then  the  chord  AB  is  a side  of  the  decagon  required. 


CONSTRUCTION  PROBLEMS 


281 


Proof.  Draw  OB  and  KB. 

Then  OA  : OK=OK : KA.  Constr. 

But  AB=OK,  OA  : AB  = AB  : KA.  Ax.  8. 
Also  Z OAB  = Z KAB.  Ident. 

A OAB  and  KAB  are  similar.  Art.  327. 

.*.  Z 0 = / AUK  ( homolog . A of  similar  A). 

And  A AKB  is  isosceles,  Art.  321. 

( being  similar  to  A AOB,  which  is  isosceles). 

.'.  AB  = KB  = KO.  Ax.  1. 

Hence  ZO=/KBO.  (Why?) 

T /.  /ABO  = 2 Z 0.  Ax.  2. 

j .'.  Z OAB  = 2 Z 0.  Art.  99. 

{ 10=  ZO. 

Adding,  Z ABO  + Z OAB  + Z 0=  5 ZO. 

But  ZABO  + ZOAB  + Z0  = 2 rt.  Z . Art.  134. 

.’.  5 Z 0 = 2 rt.  A . Ax.  l. 

.*.  Z 0=i  of  2 rt.  A , or  to  of  4 rt.  A . 
arc  AB  is  iV  of  the  circumference. 


Hence,  if  the  chord  AB  be  applied  ten  times  in  suc- 
cession to  the  circumference,  a regular  decagon  will  be 

inscribed  in  the  given  circle.  Art.  418. 

Q.  E.  F, 

463.  Cor.  1.  By  joining  the  alternate  vertices  of  a 
regular  inscribed  decagon , a regular  pentagon  can  be  inscribed 
in  a given  circle. 

464.  Cor.  2.  By  bisecting  the  arcs  AB,  BG . . . and 
drawing  chords,  a regular  polygon  of  20  sides  can  be  in- 
scribed in  a given  circle;  by  repeating  the  process  a regular 
polygon  of  40,  80,  ...  5 X 271  sides  can  be  inscribed. 

How  can  a regular  polygon  of  5 X 2n  sides  be  circum- 
scribed about  a given  circle  1 


282 


BOOK  V.  PLANE  GEOMETRY 


Proposition  XVIII.  Problem 

465.  To  inscribe  a regular  polygon  of  fifteen  sides  (pen- 
tedecagon)  in  a given  circle. 


Given  the  O AK. 

To  inscribe  a regular  pentedecagon  in  the  given  O . 

Construction.  Draw  the  chord  AC  equal  to  the  radius  of 
the  given  O , and  the  chord  AB  equal  to  a side  of  a regular 
decagon  inscribed  in  the  circle.  Art.  462. 

Draw  the  chord  BC. 

Then  BC  is  a side  of  the  required  pentedecagon. 

Proof.  AC  is  a side  of  a regular  inscribed  hexagon. 

Art.  459. 

/.  arc  AC=i  of  the  circumference. 

In  like  manner  arc  AB—ft  of  the  circumference.  Art.  462. 
.\  arc  BC=\ — tV,  or  iV  of  the  circumference.  Ax.  3. 

Hence,  if  chord  BC  be  applied  fifteen  times  in  succes- 
sion to  the  circumference,  a regular  pentedecagon  will  be 

inscribed  in  the  given  circle.  Art.  418. 

Q.  E.  F. 

466.  Cor.  Bij  bisecting  the  arcs  BC,  CD,  . . . eic.,r 
drawing  chords,  and  repeating  the  process,  regular  polygons 
of  30,  60,  . . . 15  X 2n  sides  can  be  inscribed  in  a given 
circle. 

How  can  a regular  polygon  of  15  X 2*  sides  be  circum- 
scribed about  a given  circle  1 


COMPUTATION  PROBLEMS 


283 


COMPUTATION  PROBLEMS 

Proposition  XIX.  Problem 

467.  Given  the  side  and  radius  of  a regular  inscribed 
polygon,  to  find  the  side  of  a regular  inscribed  poly gon  of  double 
the  number  of  sides,  in  terms  of  the  given  quantities . 


Given  the  circle  0 with  radius  B,  AB  a side  of  a regular 
inscribed  polygon,  and  AG  a side  of  the  regular  inscribed 


polygon  of  double  the  number  of  sides. 

To  determine  AG  in  terms  of  AB  and  R. 

Solution.  Draw  the  radii  OA  and  0(7. 

Then  OC  is  the  _L  bisector  of  AB.  (Why?) 

But  arc  AGB  < semicircumference.  (Why?) 

AC  < a quadrant.  Ax.  10. 

.\  ZAOC  is  an  acute  Z . Art.  257. 

Hence,  in  A OAC,  AC2=  0l2+  0 03—  2 OCX  OB. 

Or  AG2  — 2B2  — 2 B X OB 

But,  in  the  rt.  A OAB,  OB2  — OA" — AB" , Art.  347. 

Or  OB2  = R2—($AB)2.  Ax.  8. 

OB  = V R?  — i Zb3  = J 1/4  B2  — Alf. 
Substituting  for  OB  its  value  thus  obtained, 

~AG2  — 2E2  — B 1 A B2  — AB2  Ax.  8. 

Or  AC=  l/R  (2  E — 1/4  B2~—  AB2), 


284 


BOOK  V.  PLANE  GEOMETRY 


468.  Special  Formulas.  If  the  radius  R,  be  taken  as  1, 


If  a side  of  the  regular  inscribed  polygon  of  n sides  be 


denoted  by  Sn,  and  R=l,  then  S‘>n  = ^// 2 — 1^4  — 


469.  To  compute  approximately  the  numerical  value  of  n. 


Given  a O whose  radius  is  1,  and  whose  circumference  is 
denoted  by  C. 

To  compute  C,  i.  e.,  2 7t,  and  hence  find  the  numerical 
value  of  7t  approximately. 

Computation.  1.  Inscribe  a regular  hexagon  in  the 
given  circle,  and  denote  its  side  by  S6. 

Then  ^6=1-  Art.  459 

/.  perimeter  of  the  inscribed  hexagon  = G. 

2.  Inscribe  a regular  polygon  of  double  the  number 
(12/)  sides. 

Then,  by  the  second  formula  of  Art.  468, 


Denoting  the  perimeter  of  a regular  inscribed  polygon 
of  n sides  by  Pn, 

Pi2  = 12  (0.51763809  -}-)  — 6.21165708. 


Proposition  XX.  Problem 


$12  = 1^2  — V 4 — 1 = 0.51763809  +. 


MAXIMA  AND  MINIMA 


285 


3.  In  the  second  formula  of  Art.  468,  let  » = 12. 

S24  = I/2  —l/4  — (0.61763809  + ) 2 = 0 . 26108238+ , etc. 
Hence  Pu  = 6.26525772. 

Computing  Sis,  P48,  etc.,  in  like  manner,  the  following 
results  are  obtained: 


Su  =1/2  — Vi  — 1 =.51763809. 

524  =1/2  — 1/4  — (.5176809  )2  =.26105238. 
-S48  =1/2  — 1/4— (.26105238)2  = .13080626. 
S96  =1/2  — 1/4  — (.13080626  f =.06543817. 
^192  = 7/2  — 1/4  — ( . 06543817  )2  = . 03272346. 
S3 84  = 1/2  — C4  — ( .03272346  f = .01636228 . 
(S768  = T/2  — C4  — ( .01636228)2  = .00818126. 


.'.  P12  =6.21165708. 
.-.  P24  = 6.26  5 2 5 7 22. 
.'.  P48  =6.27870041. 
.'.  P96  =6.28206396. 

P192  = 6. 28290510. 
.-.  P384  = 6.28311544. 
.-.  P768  = 6. 28316941. 


Bv  continuing  the  computation  it  is  found  that  the  first 
six  decimal  figures  in  the  value  of  the  perimeter  of  the  in- 
scribed polygon  remain  unchanged. 

.*.  C,  or  271  = 6.283169  approximately. 

.'.  71  = 3.14159  approximately. 


MAXIMA  AND  MINIMA 

470.  Def.  A maximum  (see  Art.  268)  is  the  great- 
est of  a group  of  magnitudes,  all  of  which  satisfy  certain 
given  conditions. 

Thus,  the  diameter  is  the  maximum  chord  which  can  he  drawn  in 
a circle. 

471.  Def.  A minimum  is  the  smallest  of  a group  of 
magnitudes,  all  of  which  satisfy  certain  given  conditions. 

Thus,  of  all  lines  which  can  be  drawn  from  a given  point  to  a given 
line  the  perpendicular  is  the  minimum. 

Certain  maxima  and  minima  have  already  been  studied,  and  we 
now  proceed  to  investigate  more  particularly  those  relating  to  regular 
polygons  and  the  circle. 


286 


BOOK  V.  PLANE  GEOMETRY 


Proposition  X5I.  Theorem 

472.  Of  all  triangles  which  have  two  sides  equal,  that 
triangle  in  which  these  sides  include  a right  angle  is  the 
maximum. 


Given  the  A ABO  and  A'B'C'  in  which  AB  — A'Bf,  CA 
= C'A' , and  Z A'  is  a rt.  Z . 

To  prove  A A'B'C'  > A ABO. 


Proof.  In  the  A ACB,  draw  CD  A AB. 

Then  CD  < CA.  (Why?) 


CD  < C'A'.  Ax.  8. 

But  the  A ABC  and  A'B'C'  have  equal  bases.  Hyp. 
these  A are  to  each  other  as  their  altitudes,  CD  and 


C'A'. 


:.  A A'B'C  > A ABC. 

{for  C’A'  > CD). 


Art.  391. 


Q.  E.  D. 


473.  Isoperimetric  figures  are  figures  having  equal 
perimeters. 


Ex.  1.  Find  the  minimum  line  that  can  he  drawn  between  two 
given  parallel  lines. 

Ex.  2.  What  is  the  largest  stick  that  can  be  placed  on  a rectangu- 
lar table  12  x 5 ft.,  and  not  have  an  end  projecting  over  a side  of 
the  table  ? 


MAXIMA  AND  MINIMA 


267 


Proposition  XXII.  Theorem 

474.  Of  all  isoperimetric  triangles  which  have  the  same 
base,  the  isosceles  triangle  is  the  maximum. 


Given  the  A ABC  and  ABD  having  the  same  base  AB 
and  equal  perimeters,  AC—CB,  and  AD  and  DB  un- 
equal. 

To  prove  A ACB  > A ADB. 

Proof.  Produce  AC  to  1 , making  CF=AC. 

Draw  FB,  and  from  D as  a center,  with  a radius  equal 
to  DB,  describe  an  arc  cutting  FB  produced  in  G. 

Draw  DO  and  AO. 

Draw  CH  and  DK  A AB;  also  CL  and  DP  A FG. 

Then  Z ABF  is  a right  Z , Art.  261. 

( for  it  may  be  inscribed  in  a semicircle  whose  center  is  C and 
whose  diameter  is  ACF). 

Also  ADG  is  not  a straight  line, 

{for,  if  it  were,  the  A DAB  and  DBA  would  be  complements  of  the  = A 
DGB  and  DBG,  respectively , and  hence  would  be  equal,  and  .'.  A DAB 
would  be  isosceles,  which  is  contrary  to  the  hypothesis) . 

/.  AF=AC+  CB  = AD  + DB  — AD  DO.  Constr.  Hyp. 


But  AD-\r  DO  > AG  (Art,  92).  .*.  AF  > AG.  Ax.  8. 

.*.  BF  > BG.  Art.  111. 

£ BF  > i BG,  or  CH  > KD.  Ax.  10. 

;.  A 4C5  > A ADB.  Art.  391. 


Q.  £.  D. 


288  BOOK  V.  PLANE  GEOMETRY 

Proposition  XXIII.  Theorem 

475.  Of  isoperimetric  polygons  having  the  same  number 
of  sides,  the  maximum  is  equilateral. 


Given  ABODE  the  maximum  of  all  polygons  having  a 
given  perimeter,  and  a given  number  of  sides. 

To  prove  ABODE  equilateral. 

Proof.  If  ABODE  is  not  equilateral,  at  least  two  of  its 
sides,  as  AB  and  BO,  must  be  unequal, 

If  this  is  possible,  on  the  diagonal  AC  as  a base,  con- 
struct a triangle  having  the  same  perimeter  as  ABC,  and 
having  side  AB'  — B'C. 

Then  A AB'G  > A ABC.  Art.  474. 

To  each  of  these  unequals  add  the  polygon  ACDE. 

:.  AB'CDE  > ABODE.  Ax.  i>. 

But  this  is  contrary  to  the  hypothesis  that  ABODE  is 
the  maximum  of  the  class  of  polygons  considered. 

Hence  AB  — BC,  and  ABODE  is  equilateral. 

Q.  E.  D. 


Ex.  Of  all  circles  which  are  described  on  a given  line  as  chord, 
which  is  the  minimum  ? 


MAXIMA  AND  MINIMA 


289 


Proposition  XXIV.  Theorem 

476.  Of  all  polygons  having  all  sides  given  hut  one,  the 
maximum  can  he  inscribed  in  a semicircle  having  the  unde ° 
•iermined  side  as  a diameter. 


Given  the  polygon  ABCDEF,  the  maximum  of  all  poly- 
gons having  the  sides  BC,  CD,  DE,  EE,  FA,  in  common, 
and  AB  undetermined. 

To  prove  that  AB  is  the  diameter  of  a semicircle  in 
which  ABCDEF  can  be  inscribed. 

Proof.  Draw  lines  from  any  vertex,  E,  to  A and  B. 

The  A BE  A must  be  the  maximum  of  all  A having  the 
sides  BE  and  EA, 

(for,  if  it  is  not,  by  increasing  or  decreasing  the  angle  BE  A,  the  A BEA 
can  be  changed  till  it  is  a maximum,  the  rest  of  figure,  BCDE  and 

EFA,  meantime  remaining  unchanged;  thus  the  area  of  the  poly  - 
gon  ABCDEF  would  be  increased,  which  is  contrary  to  the 
hypothesis  that  ABCDEF  is  a maximum) . 

Z BEA  is  a right  Z . Art  472. 

.*.  E is  on  the  semicircumference  of  which  AB  is  the 
diameter. 

In  like  manner  the  other  vertices,  C,  D and  F,  must  be 
on  the  semicircumference  which  has  AB  for  a diameter. 

Q.  E.  Do 


S 


290 


BOOK  V.  PLANE  GEOMETRY 


Proposition  XXV.  Theorem 

477.  Of  all  polygons  formed  with  the  same  given  sides 
that  which  can  he  inscribed  in  a circle  is  the  maximum. 


Given  ABODE  a polygon  which  can  be  inscribed  in  a O, 
and  A'B'G'D'E'  a polygon  which  has  the  same  sides  as 
ABODE,  but  which  cannot  be  inscribed  in  a O. 

To  prove  A BODE  > A'B'G'D'E'. 

Proof.  From  any  vertex,  A,  of  ABODE  draw  the 
diameter  AK,  and  join  K to  the  adjacent  vertices  C and  D. 
Upon  CD'  construct  the  triangle  G'K!D!  equal  to  A CDK, 


and  draw  A'K' . 

Then  area  ABOK  > area  A'B'G'K! . Art.  476. 

Also  area  AEDK  > area  A'E'D'K! . (Why?) 

Adding,  ABCKDE  > A'B'CK'D'E' . (Why?) 

But  A CKD=  A OK'D'.  Constr. 

Subtracting,  ABODE  > A'B’C'D'E'.  (Why?) 


Q.  E.  D. 

478.  Note.  It  might  happen  that  one  of  the  parts  of  the  second 
figure  formed  by  the  diameter  A'K' , as  A’B'C'E' , could  be  inscribed  in 
a semicircle,  and  .'.  = ABCK.  How,  then,  would  the  above  proof  be 
modified  ? 

479.  Cor.  Of  all  isoperimetric  polygons  of  a given 
number  of  sides  the  maximum  polygon  is  regular. 

For  it  is  equilateral  (Art.  475),  and  can  be  inscribed  in 
& circle  (Art.  477),  and  is,  therefore,  regular  (Art.  417). 


maxima  and  minima 


•291 


Proposition  XX YI.  Theorem 

480.  Of  tivo  isoperimetric  regular  polygons , that  which 
has  the  greater  number  of  sides  has  the  greater  area. 


k' 


Given  K a regular  polygon  of  any  number  of  sides,  as 
three,  and  K'  a regular  polygon  of  one  more,  or  four  sides, 
and  let  K and  K'  have  equal  perimeters. 

To  prove  K 7 > K. 

Proof.  From  any  vertex,  C,  of  K,  draw  a line  CD  to  any 
point  D of  the  side  AB,  which  meets  one  of  the  sides  of 
Z C. 

Construct  the  A DCF,  having  CF—DA,  and  DF=CA. 

A DCF—  A CD  A.  Art.  101. 

Adding  A CBD,  DFCB  - K.  As.  2. 

Hence  the  polygon  DFCB  has  the  same  perimeter  as  K' , 
and  the  same  area  as  K. 

But  DFCB  is  an  irregular,  while  K ' is  a regular  poly- 
gon of  four  sides. 

.’.  K'  > DFCB.  Art.  479. 

.'.  K > K.  Ax.  8. 

In  like  manner  it  may  be  shown  that  a regular  polygon 
of  one  more,  or  five  sides,  >K,  and  so  on. 

Q.  E.  D. 

481.  Cor.  Of  isoperimetric  plane  figures,  the  circle  is 
the  maximum.  That  is,  the  area  of  a circle  is  greater  than 
the  area  of  any  polygon  with  equal  perimeter. 


292 


BOOK  Y.  PLANE  GEOMETRY 


Proposition  XXVII.  Theorem 

482.  Of  two  equivalent  regular  polygons  that  which  has 
the  less  number  of  sides  has  the  greater  perimeter. 


Given  K and  K'  two  regular  polygons  having  the  same 
area,  and  having  their  perimeters  denoted  by  P and  P',  but 
J57  having  the  less  number  of  sides. 

To  prove  P’  > P. 

Proof.  Let  H be  a regular  polygon  having  the  same 
number  of  sides  as  K',  and  the  same  perimeter  as  K. 

Then  K>  H.  Art.  480. 


Ex.  1 . Find  the  area  of  a triangle  in  which  two  of  the  sides  are  6 
and  12  in.,  and  the  included  angle  is  90°.  Find  the  area  of  another 
triangle  having  two  sides  of  6 and  12  in.,  and  the  included  angle  60°. 

Ex.  2.  How  long  is  the  fence  about  a garden  60x40  ft.?  How 
many  square  feet  in  the  area  of  the  garden  ? Find  also  the  length  of 
fence  and  area  of  a garden  50  ft.  square. 

Ex.  3.  Find  the  area  of  an  equilateral  triangle,  a square,  a regular 
hexagon,  and  a circle,  in  each  of  which  the  perimeter  is  1 ft. 

Ex.  4.  Find  the  perimeter  of  an  equilateral  triangle,  a square,  and 
a circle,  in  each  of  which  the  area  is  24  sq.  in. 

Ex.  5.  What  principle  of  maxima  and  minima  is  illustrated  in 
each  of  the  four  preceding  Exs.  ? 


K>  > IP 
P’  > P. 


Art.  399. 

Q.  E.  D. 


Ax.  8. 


Symmetry 


293 


SYMMETRY 

483.  Symmetry  of  polygons.  Many  of  the  properties 
of  regular  figures  can  be  obtained  in  a simple  and  expedi- 
tious way  by  the  use  of  the  ideas  of  symmetry. 

484.  An  axis  of  symmetry  is  a line  such  that,  if  part  of 
1 a figure  be  folded  over  upon  it  as  an  axis,  the  part  folded 

over  will  coincide  with  the  remaining  part  of  the  figure. 

EXERCISES.  CROUP  46 

Ex.  1.  How  many  axes  of  symmetry  has  an 
isosceles  triangle  f an  equilateral  triangle  ? 

Ex.  2.  How  many  has  a square  ? a regular 
pentagon  ? 

Ex.  3.  How  many  has  a regular  hexagon  ? a 
regular  heptagon  ? 

Ex.  4.  How  many  has  a regular  octagon  ? a 
regular  polygon  of  n sides  ? 

Ex.  5.  How  many  has  a circle  ? 

485.  A center  of  symmetry  for  a polygon  is  a point 
such  that  any  line  drawn  through  the  point  and  terminated 
by  the  perimeter  is  bisected  by  the  point. 

EXERCISES.  CROUP  47 

Ex.  1.  Has  an  equilateral  triangle  a center  of  symmetry  ? Has  a 
'square  ? 

Ex.  2.  Has  a regular  pentagon  a center  of  symmetry  ? Has  a 
regular  hexagon  ? 

Ex.  3.  In  general,  which  regular  polygons  have  a center  of  sym- 
metry, and  which  do  not  ? 

Ex.  4.  Has  a circle  a center  of  symmetry  f 

Ex.  5.  Which  is  the  most  symmetrical  plane  figure  studied  thus  far  f 


294 


BOOK  V.  PLANE  GEOMETRY 


SYMMETRY  WITH  RESPECT  TO  A LINE  OR  AXIS 


486.  Two  points  symmetrical  with  respect  to  a line  oi 
axis  are  points  such  that  the  straight  line  joining  fhem  is 
bisected  by  the  given  line  at  right  angles. 

Thus,  if  PP ' is  bisected  by  AB,  and  PP'  is 
X AB,  the  points  P and  P'  are  symmetrical 
with  respect  to  the  axis  AB. 

487.  A figure  symmetrical  with  respect 
to  an  axis  is  a figure' such  that  each  point 
in  the  one  part  of  the  figure  has  a point  in 
the  other  part  symmetrical  to  the  given 
point,  with  respect  to  an  axis. 

488.  Two  figures  symmetrical  with  respect 
to  an  axis  are  two  figures  such  that  each  point 
in  the  one  figure  has  a point  in  the  other 
figure  symmetrical  to  the  given  point,  with 
respect  to  an  axis. 

SYMMETRY  WITH  RESPECT  TO  A POINT 


489.  Two  points  symmetrical  with  respect  to 
a point  or  center  are  points  such  that  the 
straight  line  joining  them  is  bisected  by  the 
point  or  center. 

Thus,  if  PP'  is  bisected  by  the  point  C,  C is 
a center  of  symmetry,  with  respect  to  P and  P' . 


490.  A figure  symmetrical  with  respect 
to  a point  or  center  is  a figure  such  that 
each  point  in  the  figure  has  another  point 
in  the  figure  symmetrical  to  the  given 
point  with  respect  to  the  center. 

491.  Two  figures  symmetrical  with  respect 

to  a center  are  figures  such  that  each  point  in 
one  figure  has  a point  in  the  other  figure 
symmetrical  to  it  with  respect  to  the  center. 


SYMMETRY 


295 


Proposition  XXVIII.  Theorem 

492.  If  a figure  is  symmetrical  with  respect  to  two  axes 
which  are  perpendicular  to  each  other,  it  is  symmetrical  ivith 
respect  to  their  point  of  intersection  as  a center. 


-X 


Given  the  figure  ABC . . . H symmetrical  with  respect  to 
the  two  axes  XX'  and  YY ; and  XX'  _L  YY'  and  intersect- 
ing it  at  0. 

To  prove  ABC . . . H symmetrical  with  respect  to  0. 

Proof.  Take  any  point  P in  the  perimeter  of  the 
figure,  and  determine  the  2ioints  P'  and  P",  symmetrical  with 
respect  to  P,  by  drawing  PKP’  _L  XX',  and  PLP " _L  YY' . 


Draw  KL,  P'0,  OP" . 

Then 

PP  ||  YY,  and  PP"  ||  XX'. 

Art.  121. 

But 

PE—EP'. 

Art.  486. 

Also 

PK=  and  ||  LO. 

Art.  157. 

KP'  = and  ||  LO. 

Ax.  1 

EL  OP' is  a £17  , andXL  = and  ||  P'0. 

Art.  160. 

In  like  manner  it  may  be  shown  that  XL  = and  ||  OP". 

P'0  = and  II  OP".  As.  1,  Art.  122. 


:.  POP"  is  a straight  line  bisected  by  point  0. 

Geom.  Ax.  3. 

Hence  any  straight  line  drawn  through  0,  and  terminated 
by  the  perimeter,  is  bisected  at  0, 

{for  P is  any  point  on  the  perimeter ) . 

O is  a center  of  symmetry  for  the  given  figure.  Art.  490. 

Q.  E.  D, 


296 


BOOK  V.  PLANE  GEOMETRY 


EXERCISES.  CROUP  48 


THEOREMS  CONCERNING  REGULAR  POLYGONS  AND 
THE  CIRCLE 


Ex.  1.  The  diagonals  of  a regular  pentagon  are  equal. 

Ex.  2.  If  ABODE  is  a regular  pentagon,  the 
triangles  ABK  and  ABC  are  similar. 

Ex.  3.  In  the  same  figure  BC=KC. 

Ex.  4.  Also  KOBE  is  a parallelogram. 

Ex.  5.  Also  AC=AB  + BK. 

Ex.  6.  The  diagonals  of  a regular  pentagon  divide  each  other  in 
extreme  and  mean  ratio. 


Ex.  7.  The  apothem  of  a regular  inseribed  tri- 
angle equals  one-half  the  radius  (r=iR). 

Ex.  8.  The  altitude  of  an  equilateral  triangle 
equals  one  and  a half  times  the  radius  of  the  cir- 
cumscribed circle. 

Ex.  9.  The  altitude  of  an  equilateral  triangle  is 
the  circumscribed  circle  as  3 : 4. 


Ex.  10.  The  side  of  an  equilateral  triangle  equals  R\/ 3,  the  radius 
of  the  circumscribed  circle  being  denoted  by  R. 

Ex.  11.  The  apothem  of  a regular  inscribed  hexagon  equals  one- 

a 

half  the  side  of  a regular  inscribed  triangle  (=— r/3). 

A 

Ex.  12.  The  side  of  a regular  circumscribed  triangle  is  double  the 
side  of  the  regular  inscribed  triangle. 

Ex.  13.  Find  the  side  of  an  inseribed  square  in  terms  of  I?;  also 
the  side  of  a circumscribed  square. 

Ex.  14.  Find  the  apothem  of  an  inscribed  square,  and! also  of  a 
circumscribed  square,  in  terms  of  R. 

Ex.  15.  Find  the  areas  of  the  inscribed  and  circumscribed  squares 
in  terms  of  R,  and  show  that  one  of  these  is  double  the  other. 

q /q  t>  2 

Ex,  1 6.  Show  that  the  area  of  a regular  inscribed  triangle  is  1 ■ — • 


EXERCISES.  THEOREMS 


297 


Ex.  17.  Show  that  the  area  of  a regular  inscribed  hexagon  is 
What,  then,  is  the  ratio  of  the  area  of  a regular  inscribed  tri- 

a 

angle  to  that  of  a regular  inscribed  hexagon  ? 

Ex.  18.  The  area  of  a regular  inscribed  hexagon  is  three-fourths 
the  area  of  the  regular  circumscribed  hexagon. 

Ex.  19.  The  area  of  a regular  inscribed  hexagon  is  a mean  pro 
portional  between  the  areas  of  a regular  inscribed  and  a regular  cir 
cumscribed  triangle. 

I [Sttg.  On  a figure  similar  to  that  of  Prop.  IV,  p.  266,  let  OP  inter 
sect  AB  in  K,  and  compare  the  A OKA,  OAP,  and  OPA/.] 

Ex.  20.  The  area  of  a regular  inscribed  polygon  of  2 n sides  is  a 
mean  proportional  between  the  areas  of  regular  inscribed  and  circum- 
scribed polygons  of  n sides. 

Ex.  21.  The  area  of  a regular  inscribed  octagon  equals  the  area  of 
the  rectangle  whose  base  and  altitude  are  the  sides  of  the  circuin 
scribed  and  inscribed  squares  respectively. 

Ex.  22.  The  area  of  an  inscribed  regular  dodecagon  equals  three 
times  the  square  of  the  radius. 

Ex.  23.  An  angle  of  a regular  polygon  is  the  supplement  of  the 
angle  at  the  center. 

Ex.  24.  Diagonals  drawn  from  a vertex  of  a regular  polygon  of  n 
sides  divide  the  angle  at  that  vertex  into  n — -2  equal  parts. 

Ex.  25.  The  diagonals  formed  by  joining  the  alternate  vertices  of 
a regular  hexagon  form  another  regular  hexagon.  Find  also  the  ratio 
of  the  areas  of  the  two  hexagons. 

Ex.  26.  If  squares  be  erected  on  the  sides  of  a regular  hexagon 
the  lines  joining  their  exterior  vertices  form  a regular  dodecagon 
Find  also  the  area  of  this  dodecagon  in  terms  of  b,  a side  of  the  hexagon 

Ex.  27.  The  square  of  a side  of  an  inscribed  equilateral  triangle 
equals  the  square  of  a side  of  an  inscribed  square  added  to  the  square 
of  a side  of  an  inscribed  regular  hexagon. 

Ex.  28.  The  area  of  a circle  equals  four  times  the  area  of  a circle 
described  on  its  radius  as  a diameter, 


298 


BOOK  V.  PLANE  GEOMETKY 


Ex.  29.  The  area  of  a circular  ring  equals  the  area  of  a circle 
whose  diameter  is  the  chord  of  the  outer  circle  tangent  to  the  inner  circle- 

Ex.  30.  Given  AaBbC,  AcB,  BdC,  semi- 
circles ; prove  that  the  sum  of  the  two  cres- 
cents AcBa  and  BdCb  equals  the  area  of  the 
right  triangle  ABC. 

Ex.  31.  An  equiangular  polygon  in- 
scribed in  a circle  is  regular  if  the  number  of  its  sides  be  odd. 

[Sug.  In  the  figure  to  Prop.  Ill,  p.  264,  arc  AEDC= arc  EDCB 

arc  AE=  arc  BC,  side  AE=side  BC,  etc.] 

EXERCISES.  CROUP  49 

MAXIMA  AND  MINIMA 

Ex.  1.  Of  isoperimetric  parallelograms  with  the  same  base  the 
rectangle  is  the  maximum. 

Ex.  2.  Of  equivalent  parallelograms  with  the  same  base,  which 
has  the  minimum  perimeter  ? 

Ex.  3.  Of  isoperimetric  rectangles  which  is  the  maximum  ? 

Ex.  4.  Divide  a given  line  into  two  parts  such  that  their  product 
is  a maximum. 

Ex.  5.  Find  a point  in  the  hypotenuse  of  a right  triangle  such  that 
the  sum  of  the  squares  of  the  perpendiculars  drawn  from  the  point  to 
the  legs  shall  be  a minimum. 

Ex.  6.  How  shall  a mile  of  wire  fence  be  stretched  so  as  to  con- 
tain the  maximum  area  ? 

Ex.  7.  Find  the  area  in  acres  included  by  a mile  of  wire  fence  if 
it  be  stretched  as  a square,  a regular  hexagon,  and  a circle  respectively. 

Ex.  8.  Of  all  triangles  with  the  same  base  and  equal  altitudes,  the 
isosceles  triangle  has  the  least  perimeter. 

Ex.  9.  Of  all  polygons  of  a given  number  of  sides  inscribed  in  a 
given  circle,  the  maximum  is  regular. 

[Sug.  Prove  the  maximum  polygon  (1)  equilateral,  (2)  equi- 
angular,] 


EXERCISES.  SYMMETRY 


299 


Ex.  10.  Find  the  maximum  rectangle  inscribed  in  a circle. 

Ex.  11.  Find  the  maximum  rectangle  that  can  be  inscribed  in  a 
semicircle. 

[Sug.  Inscribe  a square  in  the  circle.] 

Ex.  12.  Of  trapezoids  inscribed  in  a semicircle  (having  the  diam 
eter  as  one  base),  find  the  maximum. 

Ex.  13.  Divide  a given  straight  line  into  two  parts  such  that  the 
eum  of  the  squares  of  these  parts  shall  be  a minimum. 


EXERCISES.  CROUP  SO 

SYMMETRY 

Ex.  1.  A rhombus  has  how  many  axes  of  symmetry  ? Has  it  a 
center  of  symmetry  ? 

Ex.  2.  What  axis  of  symmetry  has  a quadrilateral  which  has  two 
pairs  of  equal  adjacent  sides  ? Has  such  a figure  a center  of  symmetry  ? 

Ex.  3.  A parallelogram  is  symmetrical  with  respect  to  the  point 
of  intersection  of  its  diagonals. 

Ex.  4.  A segment  of  a circle  is  symmetrical  with  respect  to  what 
axis? 

Ex.  5.  Has  a trapezium  a center  of  symmetry  ? An  axis  of 
symmetry  ? 

Ex.  6.  How  many  axes  of  symmetry  have  two  equal  circles  taken 
as  one  figure  ? Have  they  a center  of  symmetry  ? 

Ex.  7.  What  axis  of  symmetry  have  any  two  circles  ? 

Ex.  8.  How  must  two  equilateral  triangles  be  placed  so  as  to  have 
a center  of  symmetry  ? So  as  to  have  an  axis  of  symmetry  ? 

Ex.  9.  If  two  polygons  are  symmetrical  with  reference  to  a cen- 
ter, any  two  homologous  sides  are  equal  and  parallel  and  drawn  in 
opposite  directions. 


300 


BOOK  Y.  PLANE  GEOMETRY 


EXERCISES.  CROUP  SI 

VARIOUS  THEOREMS 

Ex.  1.  The  square  on  the  diameter  of  a circle  equals  twice  the 
square  inscribed  in  the  circle. 

Ex.  2.  The  altitude  of  an  inscribed  equilateral  triangle  is  to  the 
radius  of  the  circle  as  3 : 2. 

Ex.  3.  The  diagonal  joining  any  two  opposite  vertices  of  a regular 
hexagon  passes  through  the  center. 

Ex.  4.  The  radius  of  an  inscribed  regular  polygon  is  a mean  pro- 
portional between  its  apothem  and  the  radius  of  the  circumscribed 
regular  polygon  of  the  same  number  of  sides. 

Ex.  5.  If  the  sides  of  a regular  hexagon  be  produced,  their  points 
of  intersection  are  the  vertices  of  another  regular  hexagon.  Also  find 
the  ratio  of  the  areas  of  the  two  hexagons. 

Ex.  6.  Of  all  lines  drawn  through  a given  point  within  an  angle, 
and  terminated  by  the  sides  of  the  angle,  the  line  which  is  bisected  at 
the  given  point  cuts  off  the  minimum  area. 

Ex.  7.  Each  angle  of  a regular  polygon  of  n -j-  2 sides  contains 

--  --  right  angles, 
n - 1-  2 

Ex.  8.  The  diagonals  from  a vertex  of  a regular  polygon  of  n + 2 
sides  divide  the  angle  at  that  vertex  into  n equal  parts. 

Ex.  9.  The  sum  of  the  perpendiculars  drawn  to  the  sides  of  a. 
regular  polygon  of  n sides  from  any  point  within  the  polygon  equals  n 
times  the  apothem. 

Ex.  10.  An  equiangular  polygon  circumscribed  about  a circle  is 
regular. 

[St/G.  Draw  radii  from  the  points  of  contact,  and  lines  from  the 
vertices  of  the  polygon  to  the  center.] 

Ex.  11.  An  equilateral  polygon  circumscribed  about  a circle  is 
regular  if  the  number  of  its  sides  is  odd. 

[Sug.  See  Figure  of  Prop.  Ill  p.  264.  Prove  AT-BQ.  Draw  radii 
and  prove  Z T=  Z Qt  etc.] 


EXERCISES.  MISCELLANEOUS  THEOREMS  301 


Ex.  12.  How  must  two  equal  isosceles  triangles  be  placed  so  as 
to  have  a center  of  symmetry  ? How  must  they  be  placed  so  as  to 
have  an  axis  of  symmetry  ? 

If,  in  a regular  inscribed  polygon  of  n sides,  Sn  denotes  a side  and  r» 
denotes  the  apothem,  show  that 

Ex.  13.  For  the  triangle,  /S3=Ei/3,  r3=iE. 

Ex.  14.  For  the  square,  Si  =R-\/2,  ri  = lR\/2. 

Ex.  15.  For  the  hexagon,  S6—R,  r6  = ^R-\/i. 

_ „ . c R(V5  — 1)  rV  10  + 21/5 

Ex.  16.  For  the  decagon,  iSio= , r 10  = * 

„ ^ x „ Rl/lO  — 2t/5,  R(V  5 + 1) 

Ex.  17.  For  the  pentagon,  S3  = — ^5 ■■■■  -• 

Ex.  18.  For  the  octagon,  £8=-E'l//2 — 1/2. 

Ex.  19.  For  the  dodecagon,  £12  = R' j/2 — -j/3. 

Ex.  20.  Prove  that  S$=R'1  + S102. 

Ex.  21.  Prove  that  S32  = S62  -f-  Sio2. 

Ex.  22.  If  ADB,  AaC,  CbB  are  semicir- 
cles and  DC  X AB,  prove  that  the  area 
bounded  by  the  three  semicircumferences 
equals  the  area  described  on  DC  as  a diameter. 


Ex.  23.  If  p„  denotes  the  perimeter  of  an  inscribed  polygon  of  n 

sides  and  Pn  the  perimeter  of  a circumscribed  polygon  of  n sides, 

xl_  r,  2 Pn  Pn 

prove  that  PSn  =-  . „ - • 

$n  A- P* 


302 


BOOK  V.  PLANE  GEOMETRI 


EXERCISES.  CROUP  62 

PROBLEMS 

Circumscribe  about  a given  circle 

Ex.  1 . An  equilateral  triangle. 

fix.  2.  A square. 

fix.  3.  A regular  pentagon.  „ 

Ex.  4.  A regular  hexagon. 

Ex.  5.  A regular  octagon. 

Ex.  6.  A regular  decagon. 

Ex.  7.  Construct  a regular  pentagram,  or  5-pointed  star. 

Ex.  8.  Construct  a hexagram,  or  6-pointed  star. 

Ex.  9.  Construct  an  8 -pointed  star. 

Ex.  10.  Construct  an  angle  of  36°. 

Ex.  11.  Construct  angles  of  18°,  9°,  72°. 

Ex.  12.  Construct  angles  of  24°,  12°,  6°,  48°. 

Ex.  13.  Divide  a given  circumference  into  two  parts  which  shall 
be  in  the  ratio  of  3 : 7. 

Ex.  14.  Construct  a regular  pentagon  which  shall  have  twice  the 
perimeter  of  a given  regular  pentagon. 

Ex.  15.  Construct  a regular  pentagon  whose  perimeter  shall  equal 
the  sum  of  the  perimeters  of  two  given  regular  pentagons. 

Ex.  16.  Construct  a regular  pentagon  whose  area  shall  be  twice 
the  area  of  a given  regular  pentagon. 

Ex.  17.  Construct  a regular  pentagon  whose  area  shall  be  equal 
to  the  sum  of  the  areas  of  two  given  regular  pentagons. 

Ex.  18.  Construct  a circumference  which  shall  be  twice  the  cir- 
cumference of  a given  circle. 


EXEECISES.  PEOBLEMS 


303 


Ex.  19.  Construct  a circle  whose  area  shall  be  three  times  the 
area  of  a given  circle. 

Ex.  20.  Construct  a circumference  equivalent  to  the  sum,  and 
another  equivalent  to  the  difference,  of  two  given  circumferences. 

Ex.  21.  Construct  a circle  equivalent  to  the  sum,  and  another 
equivalent  to  the  difference,  of  two  given  circles. 

Ex.  22.  Construct  a circle  whose  area  shall  be  two-thirds  the  area 
of  a given  circle. 

Ex.  23.  Bisect  the  area  of  a given  circle  by  a concentric  circum- 
ference. 

Ex.  24.  Divide  the  area  of  a given  circle  into  five  equal  parts  by 
drawing  concentric  circumferences. 


On  a given  line  construct 

Ex.  25.  A regular  pentagon. 

Ex.  26.  A regular  hexagon. 

Ex.  27.  A regular  dodecagon. 

Ex.  28.  A circle  equivalent  to  a given  semicircle. 

Ex.  29.  Inscribe  a regular  octagon  in  a given  square. 

Ex.  30.  Inscribe  a circle  in  a given  sector. 

Ex.  31.  Inscribe  a square  in  a given  segment. 

# 

Ex.  32.  In  a given  equilateral  triangle  inscribe  three  equal  cireles; 
each  of  which  touches  the  other  two  circles  and  a side  of  the  triangle. 

Ex.  33.  In  a given  circle  inscribe  three  equal  circles  which  shall 
touch  each  other  and  the  given  circumference. 


NUMERICAL  APPLICATIONS  OF  PLANE 
GEOMETRY 


METHODS  OF  NUMERICAL  COMPUTATIONS 

493.  Cancellation.  In  numerical  work  in  geometry,  as 
elsewhere,  the  labor  of  computations  may  frequently  be 
economized.  Those  methods  of  abbreviating  work,  which 
are  particularly  applicable  in  the  ordinary  numerical  appli- 
cations of  geometry,  maybe  briefly  indicated,  as  follows: 

To  simplify  numerical  work  by  cancellation,  group 
together  as  a whole  all  the  numerical  processes  of  a given 
problem,  and  make  all  possible  cancellations  before  proceed - 
ing  to  a final  numerical  reduction. 

Ex.  Find  the  ratio  of  the  area  of  a rectangle,  whose  base  and  alti- 
tude are  42  and  24  inches,  to  the  area  of  a trapezoid,  whose  bases  are 
21  and  35  and  altitude  12. 

By  Arts.  383,  394  3 

C t 

area  of  rectangle  _ 42X24  X ,2,4 

area  of  trapezoid  6(21  + 35)  ^X^@  ’ ai0‘ 

? 

J 

494.  Use  of  radicals  and  of  7t.  Where  radicals  enter  in 
.the  course  of  the  solution  of  a numerical  problem,  it  fre- 
quently saves  labor  not  to  extract  the  root  of  the  radical  till 

the  final  answer  is  to  be  obtained. 

Ex.  1.  Find  the  area  of  a circle  circumscribed  about  a square 
whose  side  is  8. 

The  diagonal  of  the  square  must  be  8V2  (Art.  346). 
the  radius  of  O =4y  2. 

by  Art.  449,  area  of  O = 7r(4i/2)2=327r=100.6,  Area. 


(304) 


NUMERICAL  COMPUTATIONS 


305 


Similarly  in  the  use  of  7 1,  it  frequently  saves  labor  not 
to  substitute  its  numerical  value  for  n till  late  in  the  process 
of  solution. 

Ex.  2.  Find  the  radius  of  a circle  whose  area  is  equal  to  the  sum 
1 the  areas  of  two  circles  whose  radii  are  6 and  8 inches,  respectively. 

Denote  the  radius  of  the  required  circle  by  x. 

Then,  by  Art.  449,  ■kx'1  = 36  7r  + 64  ir- 

7r  x2  = 100  7 r. 

x1  — 100,  and  10,  Radius. 

495.  Use  of  x,  y,  etc.,  as  symbols  for  unknown  quan- 
tities. Xu  some  cases  a numerical  computation  is  greatly 
facilitated  by  the  use  of  a specific  symbol  for  an  unknown 
quantity. 

Ex.  In  a triangle  whose  sides  are  12,  18,  and  25,  find  the  segments 
of  the  side  25  made  by  the  bisector  of  the  angle  opposite. 

Denote  the  required  segments  of  side  25  by 
X and  25  — x. 

Then  12  : 18=a:  : 25  — ®(Art.  332) 

.*.  18a; =12  (25— x)  (Art.  302) 

.'.  10 . d _ v. 

And  2z-x=r0)SegmentS- 

496.  Limitations  of  numerical  computations.  Owing  to 
the  limitations  of  human  eyesight  and  of  the  instruments 
used  in  making  measurements,  no  measurement  can  be 
accurate  beyond  the  fifth  or  sixth  figure;  and  in  ordinary 

tvork,  such  as  is  done  by  a carpenter,  measurements  are 
not  accurate  beyond  the  third  figure.  As  all  numerical 
applications  of  geometry  are  based  on  practical  measure- 
ments, it  is  not  necessary  to  carry  arithmetical  ivork  beyond 
the  fifth  or  sixth  significant  digit. 

Other  methods  of  facilitating  numerical  computations, 
as  by  the  use  of  logarithms,  are  beyond  the  scope  of  this 
book. 


ty  \ 

/ x \ 

25 - 

306 


PLANE  GEOMETRY 


NUMERICAL  PROPERTIES  OF  LINES 
EXERCISES.  CROUP  53 

THE  RIGHT  TRIANGLE 

Ex.  1.  Find  the  hypotenuse  of  a right  triangle  whose  legs  are  12 
and  35. 

Ex.  2.  The  hypotenuse  of  a right  triangle  is  29,  and  one  leg  is 
20.  Find  the  other  leg. 

Ex.  3.  If  a window  is  15  ft.  from  the  ground  and  the  foot  of  a 
ladder  is  to  he  8 feet  from  the  house,  how  long  a ladder  is  necessary 
to  reach  the  window  ? 

Ex.  4.  Find  the  diagonals  of  a rectangle  whose  sides  are  5 and  12. 

Ex.  5.  If  the  base  of  an  isosceles  triangle  is  8 and  a leg  is  5,  find 
the  altitude. 

Ex.  6.  Find  the  diagonal  of  a square  whose  side  is  1 ft.  6 in.* 

Ex.  7.  The  diagonals  of  a rhombus  are  24  and  10.  Find  a side. 

Ex.  8.  One  side  of  a rhombus  is  17  and  one  diagonal  is  30.  Find 
the  other  diagonal. 

Ex.  9.  In  a circle  whose  radius  is  5,  find  the  length  of  the  longest 
and  shortest  chords  through  a point  at  a distance  3 from  the  center. 

Ex.  10.  In  a circle  whose  radius  is  25  in.,  find  the  distance  from 
the  center  to  a chord  48  in.  long. 

Ex.  11.  If  a chord  12  in.  long  is  5 in.  from  the  center  of  a circlef 
how  far  from  the  center  of  the  same  circle  is  a chord  10  in.  long  f 

Ex.  12.  A ladder  40  ft.  long  reaches  a window  20  ft.  high  on  one 
side  of  a street  and,  if  turned  on  its  foot,  reaches  a window  30  ft.  high 
on  the  other  side.  How  wide  is  the  street  ? 

Ex.  13.  If  one  leg  of  a right  triangle  is  10  and  the  hypotenuse  is 
twice  the  other  leg,  find  the  hypotenuse. 

Ex.  14.  Find  the  altitude  of  an  equilateral  triangle  whose  side  is  6. 

Ex.  15.  Find  the  side  of  an  ecuilateral  triangle  whose  altitude  is  8. 


NUMERICAL,  EXERCISES.  LINES 


307 


Ex.  16.  Find  the  side  of  a square  whose  diagonal  is  15. 

Ex.  17.  One  leg  of  a right  triangle  is  3,  and  the  sum  of  the 
hypotenuse  and  the  other  leg  is  9.  Find  the  sides. 

Ex.  18.  A tree  90  ft.  high  is  broken  off  40  ft.  from  the  ground. 
How  far  from  the  foot  of  the  tree  will  the  top  strike  ? 

Ex.  19.  The  radii  of  two  circles  are  1 and  6 in.,  and  their  centers 
are  13  in.  apart.  Find  the  length  of  the  common  external  tangent. 

Ex.  20.  The  sides  of  a triangle  are  10,  11,  12.  Find  the  length  of 
the  projection  of  the  side  whose  length  is  10,  on  the  side  12. 


EXERCISES.  CROUP  54 

TRIANGLES  IN  GENERAL 

Ex.  1.  The  sides  of  a triangle  are  12,  18  and  20.  Find  the  seg- 
ments of  the  side  20,  made  by  the  bisector  of  the  angle  opposite. 

Ex.  2.  In  the  same  triangle,  find  the  segments  of  the  side  20,  made 
by  the  bisector  of  the  exterior  angle  opposite. 

Ex.  3.  If  the  legs  of  a right  triangle  are  6 and  8,  find  the  hypote- 
nuse, the  altitude  on  the  hypotenuse,  and  the  projections  of  the  legs 
on  the  hypotenuse. 

Ex.  4.  Is  a triangle  acute,  obtuse,  or  right,  if  the  three  sides  are 
5,  12,  14;  5,  11,  12;  5,  12,  13;  4,  5,  6 t 

Ex.  5.  If  the  sides  of  a triangle  are  6,  7 and  8,  compute  the  length 
of  the  altitude  on  8. 

Ex.  6.  Also  the  length  of  the  median  on  the  same  side. 

Ex.  7.  Also  the  length  of  the  bisector  of  the  angle  opposite  the 
side  8. 

Ex  8.  If  two  sides  and  a diagonal  of  a parallelogram  are  8,  12 
and  10,  find  the  other  diagonal. 

[Sug.  Use  Art.  352.] 


308 


PLANE  GEOMETRY 


Ex.  9.  Two  sides  of  a triangle  are  10  and  18,  and  the  median  to 
the  third  side  is  12.  Find  the  third  side. 

Ex.  10.  Two  sides  of  a triangle  are  17  and  16,  and  the  altitude  on 
the  third  side  is  15.  Find  the  third  side. 

Ex.  11.  The  hypotenuse  of  a right  triangle  is  10,  and  the  altitude 
©nthe  hypotenuse  is  4.  Find  the  segments  of  the  hypotenuse  and  the 
legs. 

Ex.  12.  Find  the  three  medians,  the  three  bisectors,  and  the  three 
altitudes  of  a triangle  whose  sides  are  13,  14,  15. 


EXERCISES.  GROUP  5 5 

CIRCUMFERENCES  AND  ARCS 

USmg  If  i-'y  > 

Ex.  1.  Find  the  circumference  of  a circle  whose  radius  is  1 ft.  9 in. 

Ex  2.  Find  the  radius  of  a circle  whose  circumference  is  121  ft. 

Ex.  3.  A bicycle  wheel  28  in.  in  diameter  makes,  in  an  afternoon, 
3,000  revolutions.  How  far  does  the  bicycle  travel  ? 

Ex.  4.  What  is  the  diameter  of  a wheel  which  makes  1,400  revolu- 
tions in  going  8,800  yds.  ? 

Ex.  5.  If  the  diameter  of  a circle  is  20,  find  the  length  of  an  arc  of 
60°;  also  of  83°. 

Ex.  6.  If  the  length  of  an  arc  is  14  and  the  radius  is  6,  find  the 
number  of  degrees  in  the  arc. 

Ex.  7.  If  the  arc  of  a quadrant  is  1 ft.  in  length,  find  the  diameter. 

Ex.  8.  Two  concentric  circumferences  are  88  and  132  in.  in  length, 
respectively.  Find  the  width  of  the  circular  ring  between  them. 

Ex.  9.  If  the  year  be  taken  as  365i  da.,  and  the  earth’s  orbit  a 
circle  whose  radius  is  93,250,000  miles,  find  the  velocity  of  the  earth 
in  its  orbit  per  second, 


NUMERICAL  EXERCISES.  LINES 


309 


Find  the  radius  and  circumference  of  a circle  circumscribed  about 

Ex.  10.  A square  whose  side  is  5. 

Ex.  11.  An  equilateral  triangle  whose  side  is  4. 

Ex.  12.  A rectangle  whose  sides  are  12  and  5. 

Ex.  13.  Find  the  central  angle  of  a sector  whose  perimeter  equals 
one-half  the  circumference. 

Ex.  14.  Find  the  diameter  of  a circle  circumscribed  about  a tri- 
angle whose  sides  are  7,  15,  and  20. 

Ex.  15.  Find  the  radius  of  a circle  whose  circumference  equals 
the  perimeter  of  a square  whose  diagonal  is  10. 


EXERCISES.  CROUP  56 

CHORDS,  TANGENTS,  AND  SECANTS 

Ex.  1.  Two  intersecting  chords  of  a circle  are  11  and  14  in.,  and 
the  segments  of  the  first  chord  are  8 and  3 in.  Find  the  segments  of 
the  second  chord. 

[Sug.  Denote  the  required  segments  by  a;  and  14  — x.] 

Ex.  2.  In  a circle  whose  radius  is  12  in.,  a chord  16  in.  long  is 
passed  through  a point  9 in.  from  the  center.  Find  the  segments  of 
the  chord. 

Ex.  3.  Two  secants  drawn  from  a point  to  a circle  are  24  and  27 
m.  long.  If  the  external  segment  of  the  first  is  6 in.,  find  the  external 
.•segment  of  the  second. 

Ex.  4.  From  a given  point  a secant  whose  external  and  internal 
segments  are  9 and  16  is  drawn  to  a circle.  Find  the  length  of  the 
tangent  drawn  from  the  same  point  to  the  circle. 

Ex.  5.  From  a given  point  a tangent  24  in.  long  is  drawn  to  a 
circle  whose  radius  is  18  in.  Find  the  distance  of  the  point  from  the 
center. 

Ex.  6.  If  a diameter  60  in.  long  is  divided  into  5 equal  parts  by 
chords  perpendicular  to  it,  find  the  length  of  the  chords. 


310 


PLANE  GEOMETRY 


Ex.  7.  If  a mountain  3 miles  high  is  visible  150  miles  at  sea, 
what  is  the  diameter  of  the  earth  ? 

Ex.  8.  If  the  earth  is  a sphere  of  radius  4,000  miles,  how  fai 
will  the  light  of  a lighthouse  100  ft.  high  be  visible  at  sea  ? 


EXERCISES.  CROUP  5 V 

LINES  IN  SIMILAR  FIGURES 

Ex.  1.  If  the  sides  of  a triangle  are  6,  7 and  8,  and  the  shortest 
side  of  a similar  triangle  is  18,  find  the  other  sides  of  the  second 
triangle. 

Ex.  2.  If  a post  5 ft.  high  easts  a shadow  3 ft.  long,  find  the 
height  of  a steeple  which  casts  a shadow  90  ft.  long. 

Ex.  3.  In  a triangle  whose  base  is  14  and  altitude  12,  a line  is 
drawn  parallel  to  the  base  and  at  a distance  2 from  the  base.  Find 
the  length  of  the  line  thus  drawn. 

Ex.  4.  The  upper  and  lower  bases  of  a trapezoid  are  12  and  20 
and  the  altitude  is  8.  If  the  legs  are  produced  till  they  meet,  find 
the  altitude  of  each  of  the  two  triangles  thus  formed. 

Ex.  5.  If  the  upper  and  lower  bases  of  a trapezoid  are  &i  and  6s 
and  the  altitude  is  h,  find  the  altitude  of  each  of  the  triangles  formed 
by  producing  the  legs. 

Ex.  6.  If  the  perimeters  of  two  similar  polygons  are  300  and  400 
and  a side  of  the  first  is  27,  find  the  homologous  side  of  the  second 

Ex.  7.  If  the  perimeter  of  a regular  polygon  is  three  times  the 
perimeter  of  a regular  polygon  of  the  same  number  of  sides,  what  is 
the  ratio  of  their  apothems  ? 

Ex.  8.  If  the  circumferences  of  two  circles  are  600  and  400  ft., 
what  is  the  ratio  of  their  diameters  ? 

Ex.  9.  In  the  preceding  example,  if  a chord  of  the  first  circle  is 
30,  what  is  the  length  of  a chord  in  the  second  circle,  subtending 
the  same  number  of  degrees  of  arc  ? 


NUMERICAL  EXERCISES.  AREAS 


311 


COMPUTATION  OF  AREAS 
EXERCISES.  CROUP  58 

AREAS  OP  TRIANGLES 

Ex.  1.  Find  the  area  in  acres  of  a triangular  field  whose  base 
is  300  ft.  and  altitude  200  ft. 

Ex.  2.  Find  the  area  of  a triangle  whose  sides  are  10,  17,  and  21. 

Ex.  3.  Find  the  area  in  acres  of  a field  whose  sides  are  60,  70. 
and  80  chains. 

Ex.  4.  Find  the  area  in  acres  of  a triangular  field  each  of  whose 
sides  is  10  chains. 

Find  the  area  of 

Ex.  5.  An  isosceles  triangle  whose  base  is  16,  and  each  of  whose 
legs  is  34. 

Ex.  6.  An  equilateral  triangle  whose  altitude  is  8. 

Ex.  7.  A right  triangle  in  which  the  segments  of  the  hypotenuse 
made  by  the  altitude  upon  it  are  12  and  3 ; also,  in  one  in  which  the 
'segments  are  a and  b. 

Ex.  8.  An  isosceles  right  triangle  whose  hypotenuse  is  12. 

> 

Ex.  9.  A right  triangle  in  which  the  hypotenuse  is  41  and  one  leg 
is  9. 

Ex.  10.  Find  in  two  ways  the  area  of  a triangle  whose  sides  are 
6,  5,  5. 

Ex.  11.  A side  of  a given  equilateral  triangle  is  4 ft.  longer  than 
the  altitude.  Find  the  area  of  the  triangle. 

Ex.  12.  The  area  of  an  isosceles  triangle  is  144  and  a leg  is  24. 
Find  the  base. 

Ex.  13.  The  area  of  an  equilateral  triangle  is  4i/3.  Find  a side. 

Ex.  14.  The  area  of  a triangle  is  1125,  and  a :b  : c= 2 : 3 : 4,  Find 
a,  6,  c. 

Ex.  15.  The  area  of  a triangle  is  6 sq.  in.,  and  two  of  its  sides  are 
3 and  5 in.  Find  the  remaining  side. 


312 


PLANE  GEOMETRY 


EXERCISES.  CROUP  59 

AREAS  OF  OTHER  RECTILINEAR  FIGURES 
Find  the  area  of 

Ex.  1.  A parallelogram,  whose  base  is  24  ft.  6 in.  and  whose  alti- 
tude is  12  ft.  9 in. 

Ex.  2.  A trapezoid  whose  oases  are  12  and  20  in.  and  whose  alti- 
tude is  li  ft. 

Ex.  3.  A rhombus  whose  diagonals  are  9 ft.  and  2 yds. 

Ex.  4.  A quadrilateral  in  which  the  sides  AB,  BC,  CD,  DA  are  12, 
13,  14,  15  and  the  diagonal  AC  is  17. 

Ex.  5.  A quadrilateral  in  which  the  sides  are  27,  36,  30,  25  and 
the  angle  included  between  the  first  two  sides  is  a right  angle. 

Ex.  6.  A square  whose  diagonal  is  12  in. 

Ex.  7.  Find  the  number  of  boards,  each  4 yds.  long  and  6 in. 
wide,  which  are  necessary  to  cover  a floor  48  X 24  ft. 

Ex.  8.  How  many  persons  can  stand  in  a room  15  X 9 ft-,  if  each 
person  requires  27  X 18  in.  f 

f 

Ex.  9.  The  baseball  diamond  is  a square  each  side  of  which  is 
90  ft.  What  fraction  of  an  acre  is  its  area  ? 

Ex.  10.  A rectangular  garden  contains  4,524  sq.  yds.  and  is  20 
yds.  longer  than  wide.  Find  its  dimensions. 

Ex.  11.  Each  side  of  a rhombus  is  24  ft.  and  each  of  the  larger 
angles  is  double  a smaller  one.  Find  the  area. 

Ex.  12.  Find  the  area  of  a rhombus  one  of  whose  sides  is  17,  and 
one  of  whose  diagonals  is  30. 

Ex.  13.  The  area  of  a trapezoid  is  4 acres,  one  base  is  120  yds., 
and  the  altitude  is  100  yds.  Find  the  other  base. 

Ex.  14.  The  bases  of  an  isosceles  trapezoid  are  20  and  36  and  the 
legs  are  17.  Find  the  area. 


NUMERICAL  EXERCISES.  AREAS 


313 


Ex.  15.  The  base  of  a triangle  is  20  and  the  altitude  18.  Find 
the  length  of  a line  parallel  to  the  base  which  cuts  off  a trapezoid 
whose  area  is  80  sq.  ft. 

[Sug.  Denote  the  altitude  of  trapezoid  by  18  — x and  find  its 
*spper  base  by  similar  triangles.] 

Ex.  16.  The  perimeter  of  a polygon,  circumscribed  about  a circle 
(rhose  radius  is  20,  is  340.  Find  the  area  of  the  polygon. 

Ex.  17.  The  area  of  a rectangle  is  144  and  the  base  is  three  times 
the  altitude.  Find  the  dimensions. 

Ex.  18.  Find  the  area  of  a regular  hexagon  one  of  whose  sides  is  10. 

Ex.  19.  Find  the  area  of  a regular  decagon  inscribed  in  a circle 
whose  radius  is  20. 

Ex.  20.  Find  a side  of  a regular  hexagon  whose  area  is  200  sq.  in. 


EXERCISES.  CROUP  60 

AREAS  OF  CIRCULAR  FIGURES 

Ex.  1.  Find  the  area  in  acres  of  a circle  whose  radius  is  100  yds. 

Ex.  2.  Find  the  radius  in  inches  of  a circle  whose  area  is  1 sq.  yd. 

Ex.  3.  Find  the  area  of  a circle  whose  circumference  is  p. 

Ex.  4.  Find  the  radius  of  a circle  whose  area  equals  the  sum  of 
the  areas  of  two  circles  whose  radii  are  9 and  40  in. 

Ex.  5.  Find  the  radius  of  a circle  whose  area  equals  the  sum  of 
the  areas  of  three  circles  whose  radii  are  20,  28,  29. 

Ex.  6.  In  a circle  of  radius  50  find  the  area  of  a sector  of  80°. 

Ex.  7.  Also  of  a segment  of  60°;  of  a segment  of  300°;  of  a seg 
ment  of  240°. 

Ex.  8.  In  a circle  whose  radius  is  7,  the  area  of  a sector  is  45  sq.  ft. 
Find  the  number  of  degrees  in  its  angle. 

Ex.  9.  In  a circle  whose  radius  is  10,  find  the  sum  of  the  segments 
formed  by  an  inscribed  square. 


314 


PLANE  GEOMETRY 


Ex.  10.  A circular  mill-pond,  i mile  in  diameter,  contains  a circu- 
lar island,  100  yds.  in  diameter.  Find  the  water  surface  of  the  pond 
in  acres. 

Ex.  11.  The  same  pond  is  surrounded  by  a driveway  30  ft.  wide. 
Find  the  area  of  the  driveway. 

Ex.  12.  Two  tangents  to  a circle,  whose  radius  is  15,  include  an 
engle  of  60°.  Find  the  area  included  between  the  tangents  and  the 
■adii  to  the  points  of  contact. 

Ex.  13.  Find  the  length  of  the  tether  by  which  a cow  must  be 
tied,  in  order  to  graze  over  exactly  one  acre. 

Ex.  14.  Three  equal  circles  touch  each  other  externally.  Show 

7T 

that  the  area  included  between  them  is  R2(i/ 3 — — )• 

Ex.  15.  If  the  area  included  between  three  equal  circles  which 
touch  each  other  externally  is  a square  foot,  find  the  radius  of  each 
circle  in  inches. 


EXERCISES.  CROUP  6t 

AREAS  OF  SIMILAR  FIGURES 

Ex.  1.  The  homologous  sides  of  two  similar  triangles  are  3 and  5. 
Find  the  ratio  of  their  areas. 

Ex.  2.  The  homologous  sides  of  two  similar  polygons  are  4 and  7, 
and  the  area  of  the  first  polygon  is  112.  Find  the  area  of  the  second 
polygon. 

Ex.  3.  The  radius  of  a circle  is  6.  Find  the  radius  of  a circle  hav- 
ing three  times  the  area  of  the  given  circle. 

Ex.  4.  The  areas  of  two  circles  are  as  16  to  9,  and  the  radius  of  the 
first  is  8.  Find  the  radius  of  the  second. 

Ex.  5.  The  sides  of  a triangle  are  5,  6,  7.  Find  the  sides  of  a 
similar  triangle  containing  9 times  the  area  of  the  given  triangle. 

Ex.  6.  If,  in  finding  the  area  of  a circle,  a student  uses  D= 50  as 
.2  = 50,  how  will  the  area  as  computed  differ  from  the  correct  area  T 


MISCELLANEOUS  NUMERICAL  EXERCISES  315 


Ex.  7.  In  a triangle  whose  base  is  24  in.  and  altitude  is  18  in.,  the 
altitude  is  bisected  by  a line  parallel  to  the  base.  Find  the  area  of 
the  triangle  cut  off. 

Ex.  8.  In  the  triangle  of  Ex.  7,  what  part  of  the  altitude  must  be  cut  of? 
in  order  that  the  area  of  the  triangle  be  bisected  ? 

Ex.  9.  In  a circle  whose  diameter  is  30  in.,  what  are  the  diameters 
of  concentric  circumferences  which  divide  the  area  into  three  equiva- 
lent parts  ? 

Ex.  10.  If  a circle  be  constructed  on  the  radius  of  a given  circle, 
and  segments,  one  in  each  circle,  be  formed  by  a line  drawn  from  the 
point  of  contact,  find  the  ratio  of  the  segments. 


EXERCISES.  GROUP  62 

GENERAL  NUMERICAL  EXERCISES  IN  PLANE  GEOMETRY 

Ex.  1.  The  leg  of  an  isosceles  triangle  is  10  and  the  base  is  16. 
Find  the  altitude  and  the  area. 

Ex.  2.  Find  the  area  of  a triangle  whose  sides  are  25,  39,  40. 
Also  find  the  radius  of  a circle  equivalent  to  this  triangle. 

Ex.  3.  Find  the  area  of  a regular  hexagon  inscribed  in  a circle 
whose  radius  is  2. 

Ex.  4.  The  sides  of  a triangle  are  7,  8 and  9 in.  Find  the  sides  of  a 
similar  triangle  of  four  times  the  area.  Also,  of  twice  the  area. 

Ex.  5.  The  temple  of  Herod  is  said  to  have  accommodated  210,900 
people  at  one  time.  If  each  person  required  27 X 18  in.,  and  one-third, 
the  space  inside  the  temple  be  allowed  for  walls,  sanctuaries,  etc., 
what  were  the  dimensions  of  the  temple,  if  it  was  a square  ? 

Ex.  6.  If  the  sides  of  a triangle  are  12,  16  and  21,  what  are  the 
segments  of  the  side  21  made  by  the  bisector  of  the  angle  opposite  I 

Ex.  7.  The  sides  of  a quadrilateral  in  order  are  5,  5,  4,  3,  and  the 
first  two  of  these  sides  contain  an  angle  of  60°.  Find  the  area. 

Ex.  8.  Find  the  diameter  of  a wheel  which,  in  a mile,  makes  480 
revolutions. 


316 


PLANE  GEOMETEY 


Ex.  9.  The  area  of  a trapezoid  is  112  and  the  two  bases  are  12  and 
16.  Find  the  altitude. 

Find  the  radius  of  a circle  equivalent  to 

Ex.  10.  A square  whose  side  is  10. 

Ex.  11.  An  equilateral  triangle  whose  side  is  12. 

Ex.  12.  A trapezoid  whose  bases  are  16  and  18  and  altitude  9. 

Ex.  13.  A semicircle  whose  radius  is  15. 

Ex.  14.  Find  the  diameter  of  a circle  whose  area  shall  be  equiva- 
lent to  the  sum  of  two  circles  whose  diameters  are  144  and  17. 

Ex.  15.  A circle,  a square,  and  an  equilateral  triangle  each  have 
a perimeter  of  12  yds.  Find  the  area  of  each  figure. 

Ex.  16.  In  a circle  whose  area  is  400,  the  area  of  a sector  is  125. 
Find  the  angle  of  the  sector. 

Ex.  17.  How  many  acres  are  included  within  a half-mile  running 
track,  if  the  track  is  in  the  shape  of  a rectangle  twice  as  long  as 
if.  is  wide  ? 

Ex.  18.  In  a square  whose  side  is  6 in.,  find  the  area  of  the 
inscribed  and  of  the  circumscribed  circles. 

Ex.  19.  Find  the  area  of  the  circle  circumscribed  about  a rect- 
angle whose  sides  are  40  and  9. 

Ex.  20.  One  leg  of  a right  triangle  is  12,  and  the  difference  be- 
tween the  hypotenuse  and  the  other  leg  is  8.  Find  the  area. 

Ex.  21.  Find  the  area  of  an  isosceles  right  triangle  whose  hypote- 
nuse is  20  ft. 

Ex.  22.  In  a triangle  whose  sides  are  16,  18,  20,  find  the  length  of 
the  altitude,  median,  and  bisector  of  the  angle  opposite  the  longest 

side. 

Ex.  23.  A line  16  inches  long  is  divided  internally  in  the  ratio  of 
3:5;  find  the  segments.  Also  find  the  segments  when  the  line  is 
divided  externally  in  the  same  ratio. 

Ex.  24.  A line  16  inches  long  is  divided  in  extreme  and  mean 
ratio.  Find  the  segments. 


MISCELLANEOUS  NUMERICAL  EXERCISES  317 


Ex.  25.  If  a line  is  divided  in  extreme  and  mean  ratio  and  the 
smaller  segment  is  4,  find  the  whole  line. 

Ex.  26.  A triangle  whose  altitude  is  20  is  bisected  by  a line  par- 
allel to  the  base.  Into  what  segments  is  the  altitude  divided  T 

Ex.  27.  Find  the  area  of  a square  inscribed  in  a circle  whose 
radius  is  5. 

Ex.  28.  In  a circle  whose  diameter  is  20,  a chord  is  passed  through 
a,  point  at  a distance  6 from  the  center,  perpendicular  to  the  diameter 
through  that  point.  Find  the  length  of  this  chord,  and  of  the  chords 
drawn  from  its  extremities  to  the  ends  of  the  diameter. 

Ex.  29.  Each  leg  of  an  isosceles  trapezoid  is  10,  and  one  base 
exceeds  the  other  by  16.  Find  the  altitude. 

Ex.  30.  If  three  arcs,  each  of  60°  and  having  10  for  a radius,  are 
each  concave  to  the  other  two,  find  the  area  included  by  them. 

Ex.  31.  Find  the  area  of  a trapezoid  whose  legs  are  4 and  5,  and 
whose  bases  are  8 and  11. 

Ex.  32.  A square  piece  of  land  and  a circular  piece  each  contain 
1 acre.  How  many  more  feet  of  fence  does  one  require  than  the  other  ? 

Ex.  33.  If  the  base  of  a triangle  is  doubled  and  the  altitude  re- 
mains unchanged,  how  is  the  area  affected  1 If  the  altitude  is  doubled 
and  the  base  remains  unchanged  ? If  both  the  base  and  the  altitude 
are  doubled  ? 

Ex.  34.  The  side  of  an  equilateral  triangle  is  12;  find  the  area  of 
the  inscribed,  and  of  the  circumscribed  circle. 

Ex.  35.  In  a given  circle  a chord  of  60°  is  16.  Find  the  chord  of 
120°.  Also  of  30°. 

Ex.  36.  In  a given  triangle,  equivalent  to  a rectangle  whose  sides 
are  40  and  20,  the  base  is  32.  Find  the  altitude. 

Ex.  37.  Find  the  side  of  an  equilateral  triangle  equivalent  to  a 
circle  whose  diameter  is  10. 

Ex.  38.  The  area  of  a rhombus  is  156  sq.  in.,  and  one  side  is  1 ft. 
1 in.  Find  the  diagonals. 


318 


PLANE  GEOMETRY 


Ex.  39.  The  sides  of  a triangle  are  8,  10,  12.  Find  the  areas  of 
the  triangles  made  by  the  bisector  of  the  angle  opposite  the  side  12. 

Ex.  40.  In  a circle  of  area  275  sq.  ft.,  a rectangle  of  area  150  sq.  ft- 
is  inscribed.  Show  how  to  find  the  sides  of  the  rectangle. 


EXERCISES,  CROUP  63 

EXERCISES  INVOLVING  THE  METRIC  SYSTEM 

Ex.  1.  Find  the  area  of  a triangle  of  which  the  base  is  16  dm.  and 
the  altitude  80  cm. 

Ex.  2.  Find  the  area  of  a triangle  whose  sides  are  6 m.,  70  dm., 
800  cm. 

Ex.  3.  Find  the  area  in  square  meters  of  a circle  whose  radius  is 
14  dm. 

Ex.  4.  If  the  hypotenuse  of  a right  triangle  is  17  dm.  and  one  leg 
is  150  cm.,  find  the  other  leg  and  the  area. 

Ex.  5.  If  the  circumference  of  a circle  is  1 m.,  find  the  area  of 
the  circle  in  square  decimeters. 

Ex.  6.  Find  the  area  in  hectares  and  also  in  acres,  of  a circle 
whose  radius  is  100  m. 

Ex.  7.  If  the  diagonal  of  a rectangle  is  35  dm.  and  one  side  is  800 
mm.,  find  the  area  in  square  meters,  and  also  in  square  inches. 

Ex.  8.  Find  the  area  of  a trapezoid  whose  bases  are  600  cm.  and 
2 m.,  and  whose  altitude  is  80  dm. 

Ex.  9.  If  a rectangular  field  is  700  dm.  long  and  200  m.  wide,  find 
its  area  in  hectares  and  in  acres. 

Ex.  10.  In  a given  circle  two  chords,  whose  lengths  are  15  dm. 
and  13  dm.,  intersect.  If  the  segments  of  the  first  chord  are  12  dm, 
and  3 dm.,  find  the  segments  of  the  second  chord. 

Ex.  11.  Find  in  decimeters  the  radius  of  a circle  equivalent  to  a 
square  whose  side  is  1 ft.  6 in. 

Ex.  12.  Find  in  feet  the  diameter  of  a wheel  which,  in  going  10 
kilometers,  makes  5,000  revolutions. 


SOLID  GEOMETRY 

Book  YI 

LINES,  PLANES  AND  ANGLES  IN  SPACE 

DEFINITIONS  AND  FIRST  PRINCIPLES 

497.  Solid  Geometry  treats  of  the  properties  of  space  of 
three  dimensions. 

Many  of  the  properties  of  space  of  three  dimensions  are  determined 
by  use  of  the  plane  and  of  the  properties  of  plane  figures  already 
obtained  in  Plane  Geometry. 

498.  A plane  is  a surface  such  that,  if  any  two  points 
in  it  be  joined  by  a straight  line,  the  line  lies  wholly  in  the 
surface. 

499.  A plane  is  determined  by  given  points  or  lines,  if 
no  other  plane  can  pass  through  the  given  points  or  lines 
without  coinciding  with  the  given  plane. 

500.  Fundamental  property  of  a plane  in  space.  A 

plane  is  determined  by  any  three  points  not  in  a straight  line. 

For,  if  through  a line  con- 
necting two  given  points,  A 
and  B,  a plane  be  passed,  the 
plane,  if  rotated,  can  pass 
through  a third  given  point, 

C,  in  but  one  position. 

The  importance  of  the  above  principle  is  seen  from  the  fact  that  it 
reduces  an  unlimited  surface  to  three  points,  thus  making  a vast 
economy  to  the  attention.  It  also  enables  us  to  connect  different 
planes,  and  treat  of  their  properties  systematically. 

(319) 


320 


BOOK  YI . SOLID  GEOMETRY 


501.  Other  modes  of  determining  a plane.  A plane  may 
also  be  determined  by  any  equivalent  of  three  points  not  in 
a straight  line,  as  by 

a straight  line  and  a point  outside  the  line;  or  by 
two  intersecting  straight  lines;  or  by 
two  parallel  straight  lines. 

It  is  often  more  convenient  to  use  one  of  these  latter  methods  of 
determining  a plane  than  to  reduce  the  data  to  three  points  and  use 

Art.  500. 


502.  Representation  of  a plane  in  geometric  figures.  In 
reasoning  concerning  the  plane,  it  is  often  an  advantage 
to  have  the  plane  represented  in  all  directions.  Hence,  in 
drawing  a geometric  figure,  a plane  is  usually  represented 
to  the  eye  by  a small  parallelogram. 

This  is  virtually  a double  use  of  two  intersecting  lines,  or  of  two 
parallel  lines,  to  determine  a plane  (Art.  501). 

503.  Postulate  of  Solid  Geometry.  The  principle  of 
Art.  499  may  also  be  stated  as  a postulate,  thus: 

Through  any  three  points  not  in  a straight  line  (or  their 
equivalent)  a plane  may  be  passed. 

504.  The  foot  of  a line  is  the  point  in  which  the  line 
intersects  a given  plane. 

505.  A straight  line  perpendicular  to  a plane  is  a line 
perpendicular  to  every  line  in  the  plane  drawn  through  its 

foot. 

A straight  line  perpendicular  to  a plane  is  sometimes  called  a normal 
to  the  plane. 


LINES  AND  PLANES 


321 


506.  A parallel  straight  line  and  plane  are  a line  and 
plane  which  cannot  meet,  however  far  they  be  produced. 

507.  Parallel  planes  are  planes  which  cannot  meet, 
however  far  they  be  produced. 

508.  Properties  of  planes  inferred  immediately. 

1.  A straight  line,  not  in  a given  plane,  can  intersect  the 
given  plane  in  hut  one  point. 

For,  if  the  line  intersect  the  given  plane  in  two  or  more 
points,  by  definition  of  a plane,  the  line  must  lie  in  the 
plane.  Art.  498. 

2.  The  intersection  of  two  planes  is  a straight  line. 

For,  if  two  points  common  to  the  two  planes  be  joined 

by  a straight  line,  this  line  lies  in  each  plane  (Art.  498) ; 
and  no  other  point  can  be- common  to  the  two  planes,  for, 
through  a straight  line  and  a point  outside  of  it  only  one 
plane  can  be  passed.  Art.  501. 


Ex.  1.  Give  an  example  of  a plane  surface;  of  a curved  surface; 
of  a surface,  part  plane  dhd  part  curved;  of  a surface  composed  of 
different  plane  surfaces. 

Ex.  2.  Four  points,  not  all  in  the  same  plane  determine  how 
many  different  planes  ? how  many  different  straight  lines  ? 

Ex.  3.  Three  parallel  straight  lines,  not  in  the  same  plane,  deter- 
mine how  many  different  planes  ? 

Ex.  4.  Four  parallel  straight  lines  can  determine  how  many  differ- 
ent planes  1 

Ex.  5.  Two  intersecting  straight  lines  and  a point,  not  in  their 
plane,  determine  how  many  different  planes  ? 


322 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  I.  Theorem 

509.  If  a straight  line  is  perpendicular  to  each  of  two 
other  straight  lines  at  their  point  of  intersection , it  is  per- 
pendicular to  the  plane  of  those  lines. 


A 


Given  AB  J_  lines  BG  and  BD,  and  the  plane  MW  pass- 
ing through  BG  and  BD. 

To  prove  AB  _l_  plane  MW. 

Proof.  Through  B draw  BG,  any  other  line  in  the  plane 
MW. 

Draw  any  convenient  line  CD  intersecting  BC,  BG  and 
BD  in  the  points  C,  G and  D,  respectively. 

Produce  the  line  AB  to  F,  making  BF=AB. 

Connect  the  points  C,  G,  D with  A,  and  also  with  F. 


Then,  in  the  A ACD  and  FCD,  CD=CD.  Ident. 

AC=CF,  and  AD  = DF.  Art.  112. 

A ACD=  A FCD.  (Why?) 

.-.  Z ACD  = Z FCD.  (Why?) 

Then,  in  the  A ACG  and  FCG,  CG=CG,  (Why?) 

AC=CF,  and  Z ACG  = l FCG.  (Why?) 

A ACG— A FCG.  (Why?) 


LINES  AND  PLANES 


323 


AG  = GF.  (Why  f ) 

.*.  B and  G are  each  equidistant  from  the  points  A and  F. 

BG  is  _L  AF\  that  is,  AB  _L  BG.  Art.  113. 

. .*.  AB  _L  plane  MN,  Art.  505. 

(for  it  is  JL  any  line,  BG,  in  the  plane  MN,  through  its  foot). 


Proposition  II..  Theorem 

510.  All  the  perpendiculars  that  can  he  drawn  to  a 
given  line  at  a given  point  in  the  line  lie  in  a plane  perpen- 
dicular to  the  line  at  the  given  point. 


Given  the  plane  MN  and  the  line  B C both  X line  AB  at 
the  point  B. 

To  prove  that  BG  lies  in  the  plane  MN. 

Proof.  Pass  a plane  AF  through  the  intersecting  lines 
AB  and  BG.  Art.  503. 

This  plane  will  intersect  the  plane  MN  in  a straight 
line  BF.  Art.  508,  2. 

But  AB  X plane  MN  fHyp.)  .*.  AB  X BF.  Art.  505. 
Also  AB  X BG.  Hyp. 

in  the  plane  AF,  BG  and  BF  X AB  at  B. 

BG  and  BF  coincide.  Art.  71. 

But  BF  is  in  the  plane  MN. 

BG  must  be  in  the  plane  MN, 

( for  BC  coincides  with  BF , which  lies  in  the  plane  MN). 

Q.  E.  D. 


324 


BOOK  VI.  ' SOLID  GEOMETRY 


511.  Cor.  1.  At  a given  point  B in  the  straight  line  AB, 
to  construct  a plane  perpendicular  to  the  line  AB.  Pass  a 
plane  AF  through  AB  in  any  convenient  direction,  and  in 
the  plane  AF  at  the  point  B construct  BF  _L  AB  (Art. 
274).  Pass  another  plane  through  AB,  and  in  it  construct 
BP  -L  AB.  Through  the  lines 
BF  and  BP  pass  the  plane 
SM2T  (Art.  503).  MN  is  the 
required  plane  (Art.  509). 

/ 512.  Cor.  2.  Through  a 
given  external  point , P,  to  pass 
a plane  perpendicular  to  a 
given  line,  AB.  Pass  a plane  through  AB  and  P (Art. 
503),  and  in  this  plane  draw  PB  JL  AB  (Art.  273).  Pass 
another  plane  through  AB,  as  AF,  and  in  AF  draw  BF  _L 
AB  at  B (Art.  274).  Pass  a plane  through  BP  and  BF 
(Art.  503).  This  will  be  the  jMane  required  (Art.  509). 

513.  Cor.  3.  Through  a given  point  but  one  plane  can 
be  passed  perpendicular  to  a given  line. 


Ex.  1.  Five  points,  no  four  of  which  are  in  the  same  plane,  deter- 
mine how  many  different  planes  ? how  many  different  straight  lines  ? 

Ex.  2.  A straight  line  and  two  points,  not  all  of  which  are  in  the 
same  plane,  determine  how  many  different  planes  ? 

Ex.  3.  In  the  figure,  p.  322,  prove  that  the  triangles  GAD  and 
GDF  are  equal. 

Ex.  4.  In  the  same  figure,  if  AB= 8 and  BG= 6,  find  FC. 


LINES  AND  PLANES 


325 


Proposition  III.  Problem 

514.  At  a given  point  in  a plane,  to  erect  a perpendicu- 
lar to  the  plane. 


Given  the  point  A in  plane  MN. 

To  construct  a line  perpendicular  to  MN  at  the  point  A. 
Construction.  Through  the  point  A draw  any  line  CD  in 
the  plane  MN. 

Also  through  the  point  A pass  the  plane  PQ  _L  CD 
(Art.  511),  intersecting  the  plane  MN  in  the  line .5$.  Art.  508,  2. 


In  the  plane  PQ  draw  AK  JL  line  ES  at  A.  Art.  274. 
Then  AK  is  the  _L  required. 

Proof.  CD  _L  plane  PQ.  Constr. 

•.  CD  J_  AK.  Art.  505. 

Hence  AK  A CD. 

But  AK  A ES.  ' Constr. 

.*.  AK  A plane  MN.  Art.  509. 


Q.  E.  F. 

515.  Cor.  At  a given  point  in  a plane  but  one  perpen- 
dicular to  the  plane  can  be  drown.  For,  if  two  A could  be 
drawn  at  the  given  point,  a plane  could  be  passed  through 
them  intersecting  the  given  plane.  Then  the  two  A would 
be  in  the  new  plane  and  A to  the  same  line  (the  line  of 
intersection  of  the  two  planes,  Art.  505),  which  is  im- 
possible (Art.  71). 


326 


BOOK  VI.  SOLID  GEOMETKY 


Proposition  IV.  Problem 

516.  From  a given  point  without  a j)lane,  to  draw  a line 
perpendicular  to  the  plane. 

At 


Given,  the  plane  MN  and  the  point  A external  to  it. 

To  construct  from  A a line  _L  plane  MN. 

Construction.  In  the  plane  MN  draw  any  convenient 
line  BC.  Pass  a plane  through  BC  and  A (Art.  503),  and 


in  this  plane  di*aw  AD  A BC.  Art.  273. 

In  the  plane  MN  draw  LD  A BC.  Art.  274. 

Pass  a plane  through  AD  and  LD  (Art.  503),  and  in 
that  plane  draw  AL  A LD.  Art.  273. 

Then  AL  is  the  A required. 

Proof.  Take  any  point  C in  BC  except  D,  and  draw  LG 
and  AC. 

Then  A ADC,.  ADL  and  LDC  are  right  A.  Constr. 

.-.  AGl  = AD1  + DC2.  Art.  400. 

.'.  AC' = AL'  -p  LD'  -p  DC'.  Art.  400,  Ax.  8. 
.’.  AC2  = AL'  -p  LC' . Art.  400,  Ax.  8. 

.'.  ZALC  is  a right  Z . Art.  351. 

But  AL  A LD.  Constr. 

AL  A MN.  Art.  509. 

Q.  E.  F. 


517.  Cor.  But  one  perpendicular  can  he  drawn  from  a 
given  external  point  to  a given  plane. 


LINES  AND  PLANES 


327 


Proposition  V.  Theorem 

518.  I.  Oblique  lines  drawn  from  a point  to  a plane, 
meeting  the  plane  at  equal  distances  from  the  foot  of  the 
perpendicular,  are  equal; 

II.  Of  two  oblique  lines  drawn  from  a point  to  a plane, 
but  meeting  the  plane  at  unequal  distances  from  the  foot  of 
the  perpendicular , the  more  remote  is  the  greater. 


A 


Given  AB  _L  plane  MN,  BB  = BC,  and  BH  > BC. 
To  prove  A B = AC,  and  AH  > AC. 

Proof.  I.  In  the  right  A ABB  and  ABC, 


AB  — AB,  and  BB  = BC.  (Why  T) 

.'.  A ABB=  A ABC.  (Why?) 

/.  AB  = AC.  (Why?) 

II.  On  BH  take  BF=BC,  and  draw  AF. 

Then  AF—  A G ( by  part  of  theorem  just  proved ) . 

But  AH  > AF.  (Why?) 

.'.  AH  > AC.  Ax.  8. 


Q.  E.  D. 

519.  Cor.  1.  Conversely:  Equal  oblique  lines  drawn 
from  a point  to  a plane  meet  the  plane  at  equal  distances 
from  the  foot  of  the  perpendicular  draivn  from  the  same  point 
to  the  plane;  and,  of  two  unequal  lines  so  drawn,  the  greater 
line  meets  the  plane  at  the  greater  distance  from  the  foot  of 
the  perpendicular. 


328 


BOOK  YI . SOLID  GEOMETRY 


520.  Cor.  2.  The  locus  of  a point  in  space  equidistant 
from  all  the  points  in  the  circumference  of  a circle  is  a 
straight  line  passing  through  the  center  of  the  circle  and 
perpendicular  to  its  plane. 

521.  Cor.  3.  The  perpendicular  is  the  shortest  line  that 
can  he  drawn  from  a given  point  to  a given  plane. 

522.  Def.  The  distance  from  a point  to  a plane  is  the 
perpendicular  drawn  from  the  point  to  the  plane. 

Proposition  VI.  Theorem 

523.  If  from  the  foot  of  a perpendicular  to  a plane  a 
line  he  drawn  at  right  angles  to  any  line  in  the  plane , the  line 
drawn  from  the  point  of  intersection  so  formed  to  any  point  in 
the  perpendicular , is  perpendicular  to  the  line  of  the  plane. 


Given  AB  _L  plane  .307",  and  BF  _L  CD,  any  line  in  MF. 
To  prove  AF  _L  CD. 

Proof.  On  CD  take  FP  and  FQ  equal  segments. 

Draw  AP,  BP,  AQ,  BQ. 

Then  BP=BQ.  Art.  112. 

Hence  AP=AQ.  Art.  518. 

/.  in  the  line  AF,  the  point  A is  equidistant  from  P 
and  Q,  and  F is  equidistant  from  P and  Q. 

:.  AF  _L  CD.  (why?) 

Q.  E.  D. 

Ex.  In  the  above  figure,  if  AB=6,  AF=S,  and  AQ= 10,  find  QF, 
BF  and  BQ. 


329 


LINES  AND  PLANES 


Proposition  YII. 


Theorem 


524.  Two  straight  lines  perpendicular  to  the  same  plane 
are  parallel. 


Given  tlie  lines  AB  and  CD  _L  plane  MN. 

To  prove  AB  ||  CD. 

Proof.  Draw  BD,  and  through  D,  in  the  plane  MN, 


draw  FH  X BD. 

Draw  AD. 

Then  BD  X FIT.  Constr. 

AD  X FH.  Art.  523. 

CD  X FH.  Art.  505. 

.\  BD , AD  and  CD  are  all  X FH  at  the  point  D. 

.*.  BD,  AD  and  CD  all  lie  in  the  same  plane.  Art.  510. 

.'.  AB  and  CD  are  in  the  same  plane.  (Why?) 
But  AB  and  CD  are  X BD.  Art.  505. 

.'.  AB  and  CD  are  ||.  Art.  121. 


Q.  E.  D. 

525.  Cor.  1.  If  one  of  two  parallel  lines  is  perpendicu- 
lar to  a plane,  the  other  is  perpendicular  to  the  plane  also. 

For,  if  AB  and  CD  be  || , and  AB  X 
plane  PQ,  a line  drawn  from  C X PQ 
must  be  ||  AB.  Art.  524. 

But  CD  must  coincide  with  the  line 
so  drawn  (Art.  47,  3);  CD  X PQ. 


4 c 


\ i p\ 


■9 


330 


BOOK  VI.  SOLID  GEOMETKY 


526.  Cor.  2.  If  two  straight  lines  are  each  parallel  to 
a third  straight  line,  they  are  parallel  to  each  other.  For, 
if  a plane  be  drawn  _L  to  the  third  line,  each  of  the  two 
other  lines  must  be  J_  to  it  (Art.  525),  and  therefore  be  || 
to  each  other  (Art.  524). 


Proposition  VIII.  Theorem 

527.  If  a straight  line  external  to  a given  plane  is  paral- 
lel to  a line  in  the  plane,  then  the  first  line  is  parallel  to  the 
given  plane. 


Given  the  straight  line  AB  ||  line  CD  in  the  plane  MN. 
To  prove  AB  ||  plane  MN. 

Proof.  Pass  a plane  through  the  ||  lines  AB  and  CD. 

If  AB  meets  MU  it  must  meet  it  in  the  line  CD. 

But  AB  and  CD  cannot  meet,  for  they  are  ||.  Art.  120. 

AB  and  MJf  cannot  meet  and  are  parallel.  Art.  506. 

Q.  E.  D. 

528.  Cor.  1.  If  a straight  line  is  parallel  to  a plane, 
the  intersection  of  the  plane  with  any  plane  passing  through 
the  given  line  is  parallel  to  the  given  line . 


LINES  and  planes 


331 


529.  Coe.  2.  Through  a given  line  A B 

(CD)  to  pass  a plane  parallel  to  another  c 
given  line  {AB) . 

Through  P,  any  point  in  CD,  draw 
QR  ||  AB  (Art.  279) . Through  CD  and 
QR  pass  a plane  (Art.  503).  This  will  be  the  plane 
required  (Art.  527). 

If  AB  and  CD  are  not  parallel,  but  one  plane  can  be 
drawn  through  CD\\AB. 


Proposition  IX.  Theorem 


530.  Two  planes  perpendicular  to  the  same  straight  line 
are  parallel. 


Given  the  planes  MN  and  PQ  ± line  AB. 

To  prove  MN  1 1 PQ. 

Proof.  If  MN  and  PQ  are  not  parallel,  on  being  pro- 
duced they  will  meet. 

We  shall  then  have  two  planes  drawn  from  a point  per- 
pendicular to  a given  line,  which  is  impossible.  Art.  513. 

MN  and  PQ  are  parallel.  Art.  507. 

Q.  E.  D, 


332  BOOK  VI.  SOLID  GEOMETRY 

Proposition  X.  Theorem 

531.  If  two  parallel  planes  are  cut  by  a third  plane , the 
intersections  are  parallel  lines. 


Given  MN  and  PQ  two  ||  plaues  intersected  by  the  plan« 
BS  in  the  lines  AB  and  CD. 

To  prove  AB  ||  CD. 

Proof.  AB  and  CD  lie  in  the  same  plane  BS. 

Also  AB  and  CD  cannot  meet;  for  if  they  did  meet 

the  planes  MN  and  PQ  would  meet,  which  is  impossible. 

Art.  507. 

.'.  AB  and  CD  are  parallel.  Art.  41. 

Q.  E.  D. 

532.  Cor.  1.  Parallel  lines  included  between  parallel 
planes  are  equal.  For,  if  AC  and  BD  are  two  parallel  lines, 
a plane  may  be  passed  through  them  (Art.  503),  intersect- 
ing MN  and  PQ  in  the  ||  lines  AB  and  CD.  Art.  531. 

.’.  ABDC  is  a parallelogram.  Art.  147. 

.’.  AC=BD. 


Art.  155. 


LINES  AND  PLANES 


333 


533.  Cor.  2.  Two  parallel  planes  are  everywhere  equi ■> 
distant. 

For  lines  _L  to  one  of  them  are  ||  (Art.  524).  Hence 
the  segments  of  these  lines  included  between  the  ||  planes 
are  equal  (Art.  532). 


Proposition  XI.  Theorem 

534.  If  two  intersecting  lines  are  each  parallel  to  a given 
plane,  the  plane  of  these  lines  is  parallel  to  the  given  plane. 


Given  the  lines  AB  and  CD.  each  il  plane  PQ,  and  inter- 
secting in  the  point  F-,  and  MK  a plane  through  AB  and 
CD. 

To  prove  MR  \\  PQ. 

Proof.  From  the  point  F draw  FR  A PQ. 

Pass  a plane  through  FC  and  FR,  intersecting  PQ  in 
RK ; also  pass  a plane  through  FB  and  FR,  intersecting 


PQ  in  RL. 

Then  RK  ||  FC,  and  RL  ||  FB.  Art.  528. 

But  FR  A RK  and  RL.  Art.  505. 

.'.  FR  A FC  and  FB.  Art.  123. 

.*.  FR  A MN.  Art.  509. 

A MN  II  PQ.  Art.  530. 


334 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XII.  Theorem 

535.  A straight  line  perpendicular  to  one  of  two  parallel 
planes  is  perpendicular  to  the  other  also. 


Given  the  plane  MN  ||  plane  PQ,  and  AB  _L  PQ. 

To  prove  AB  A.  MN. 

Proof.  Through  AB  pass  a plane  intersecting  PQ  and 
MN  in  the  lines  BC  and  AF,  respectively;  also  through 
AB  pass  another  plane  intersecting  PQ  and  MN  in  BD  and 
AH,  respectively. 

Then  BC  II  AF,  and  BD  ||  AH.  Art.  531. 

But  AB  i_  BC  and  BD.  Art.  505. 

.*.  AB  A AF  and  AH.  Art.  123. 

.'.  AB  X plane  MN.  Art.  509. 

0.  E.  D. 

536.  Cor.  1.  Through  a given  point  to  pass  a plane 
parallel  to  a given  plane. 

Let  the  pupil  supply  the  construction. 

537.  Cor.  2.  Through  a given  point  but  one  plane  can 
be  passed  parallel  to  a given  plane, 


LINES  AND  PLANES 


335 


Proposition  XIII.  Theorem 

538.  If  two  angles  not  in  the  same  plane  have  their 
corresponding  sides  parallel  and  extending  in  the  same  direc~ 
tion , the  angles  are  equal  and  their  planes  are  parallel. 


Given  the  ABAC  in  the  plane  JOT,  and  the  AB'A'C ' 
in  the  plane  PQ;  AB  and  A'B'  ||  and  extending  in  the 
same  direction;  and  AC  and  A'C  ||  and  extending  in  the 
same  direction. 

To  prove  ABAC  = AB'A'C , and  plane  MN  ||  plane  PQ. 


Proof.  Take  AB  — A'B',  and  AG = A'C. 
Draw  AA',  BB',  CC',  BC,  B'C'. 


Then  ABB' A1  is  a OJ  , 

( for  AB  and  A'B'  are— and  ||  ). 

Art.  160. 

/.  BB'  and  A A'  are  = and  ||. 

Art.  155. 

In  like  manner  CC'  and  A A'  are  = and  ||. 

BB'  and  CC'  are  = and  ||. 

(Why  ?) 

BCC'B'  is  a CO  , and  BC=B'C'. 

(Why?) 

A ABC=  A A' B'C'. 

(Why  ?) 

AA  - AA'. 

(Why?) 

Also 

AB  ||  A'B',  :.  AB  ||  plaue  PQ. 

Art.  527. 

Similarly 

AC  ||  plane  PQ. 

plane  MN  ||  plane  PQ. 

Art.  534. 

0.  V.  D. 

336 


BOOK  VI.  SOLID  GEOMETBY 


Proposition  XIV.  Theorem 

539.  If  tivo  straight  lines  are  intersected  by  three  paral- 
lel planes,  the  corresponding  segments  of  these  lines  are 
proportional. 


Given  the  straight  lines  AB  and  CD  intersected  by  the 
||  planes  MN,  PQ  and  ES  in  the  points  A,  F,  B,  and  C 


E,  D,  respectively. 

To  prove 


AF  CH 
FB  ED 


Proof. 

Draw  the  line  AD  intersecting  the  plane  PQ  in  G. 

Draw  FG,  BD,  GE,  AC. 

Then 

FG  II  BD,  and  GE  ||  AC. 

Art.  531. 

b|!^ 

ll 

Art.  517. 

And 

CE_AG 
ED  GD 

(Why?) 

. AF _ CE 
FB  . ED ' 

(Why  ?) 

Q.  E.  D. 

Ex.  1.  In  above  figure,  if  AF=2,  FB= 5,  and  CH—  3,  find  CD. 
Ex  2.  If  CH=  3,  ED= 4,  and  AB= 10,  find  AF  and  BF. 


DIHEDRAL  ANGLES 


337 


DIHEDRAL  ANGLES 

540.  A dihedral  angle  is  the  opening  between  two  in- 
tersecting planes. 

From  certain  points  of  view,  a dihedral  angle  may  be  regarded  as 
a wedge  or  slice  of  space  cut  out  by  the  planes  forming  the  dihedral 
angle. 

541.  The  faces  of  a dihedral  angle  are  the  planes  form- 
ing the  dihedral  angle. 

The  edge  of  a dihedral  angle  is  the  straight  line  in  which 
the  faces  intersect. 

542.  Naming  dihedral  an- 
gles. A dihedral  angle  may  be 
named,  or  denoted,  by  naming 
its  edge,  as  the  dihedral  angle 
AB\  or  by  naming  four  points, 
two  on  the  edge  and  one  on 
each  face,  those  on  the  edge 
coming  between  the  points  on 
the  faces,  as  P-AB-Q.  The 
latter  method  is  necessary  in  naming  two  or  more  dihedral 
angles  which  have  a common  edge. 

543.  Equal  dihedral  angles  are  dihedral  angles  which 
can  be  made  to  coincide. 

544.  Adjacent  dihedral  angles  are  dihedral  angles  hav- 
ing the  same  edge  and  a face  between  them  in  common. 

545.  Vertical  dihedral  angles  are  two  dihedral  angles 
having  the  same  edge,  and  the  faces  of  one  the  prolonga- 
tions of  the  faces  in  the  other. 

546.  A right  dihedral  angle  is  one  of  two  equal  adja- 
cent dihedral  angles  formed  by  two  planes. 


v 


338 


BOOK  VI.  SOLID  GEOIIETBY 


547.  A plane  per- 
pendicular to  a given 
plane  is  a plane  form- 
ing a right  dihedral 
angle  with  the  given 
plane. 


Many  of  the  properties  of  dihedral  angles  are  obtained 


most  conveniently  by  using  a plane  angle  to  represent  the 
dihedral  angle. 


548.  The  plane  angle  of  a dihedral  an- 
gle is  the  angle  formed  by  two  lines  drawn 
one  in  each  face,  perpendicular  to  the  edge 
at  the  same  point. 

Thus,  in  the  dihedral  angle  C-AB-F, 
if  PQ  is  a line  in  the  face  AD  perpendicu- 
lar to  the  edge  AB  at  P,  and  PR  is  a line 
in  face  AF  perpendicular  to  the  edge  AB 
at  P,  the  angle  QPR  is  the  plane  angle 
angle  C-AB-F. 


of  the  dihedral 


549.  Property  of  plane  angles  of  a dihedral  angle. 

The  magnitude  of  the  plane  angle  of  a dihedral  angle  is 
the  same  at  every  point  of  the  edge.  For  let  EAC  be  the 
plane  Z of  the  dihedral  Z E-AB-D  at  the  point  A. 

Then  PR  ||  AE,  and  PQ  ||  AC  (Art.  121.) 

.-.  /.RPQ  = ZEAC  (Art.  538). 


550.  The  projection  of  a point  upon  a plane  is  the  foot  of 

a perpendicular  drawn  from  the  point  to  the  plane. 

A 


551.  The  projection  of  a line 
upon  a plane  is  the  locus  of  the  pro- 
jections of  all  the  points  of  the  line 
on  the  plane.  Thus  A'B'  is  the 
projection  of  AB  on  the  plane  MN. 


tv. 


DIHEDRAL  ANGLES 


339 


Proposition  XV.  Theorem 


552.  Two  dihedral  angles  are  equal  if  their  plane  angles 
are  equal. 


Given  Z DBF  the  plane  Z of  the  dihedral  Z C-AB-F , 
Z D'B'F'  the  plane  Z of  the  dihedral  Z G'-A'B'-F' , and 
Z DBF  = Z D'B'F. 

To  prove  Z C-AB-F  = Z C'-A'B'-F'. 

Proof.  Apply  the  dihedral  Z C'-A'B'-F'  to  Z C-AB-F 

so  that  Z D'B'F'  coincides  with  its  equal,  /.DBF. 

Geom.  Ax.  2. 

Then  line  A'B'  must  coincide  with  AB,  Art.  515. 

( for  A'B'  and  AB  are  both  L plane  DBF  at  the  point  B). 

Hence  the  plane  A'B'D'  will  coincide  with  plane  ABD, 

Art.  501. 

( through  two  intersecting  lines  only  one  plane  can  be  passed) . 

Also  the  plane  A'B'F'  will  coincide  with  the  plane  ABF, 

( same  reason). 

:.  Z C'-A'B'-F'  coincides  with  Z C-AB-F  and  is  equal 

to  it.  Art.  47. 

Q.  E.  D. 

553.  Cor.  The  vertical  dihedral  angles  formed  by  two 
intersecting  planes  are  equal. 

In  like  manner,  many  other  properties  of  plane  angles 
are  true  of  dihedral  angles. 


340 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XVI.  Theorem 


554.  Two  dihedral  angles  have  the  same  ratio  as  their 
plane  angles. 


Given  the  dihedral  A C-AB-D  and  C'-A'B'-D'  having 
the  plane  A CAB  and  C’A'D’ , respectively. 

To  prove  Z C'-A'B'-D ' : Z C-AB-D  = Z C'A'D'  : 
ACAD. 


Case  I.  When  the  plane  A C’A'D'  and  CAD  (Figs.  2 
and  1),  are  commensurable. 

Proof.  Find  a common  measure  of  the  A C’A’D'  and 
CAD,  as  Z CAK,  and  let  it  be  contained  in  Z C’A'D'  n 
times,  and  in  Z CAD  m times. 

Then  Z C'A'D'  : Z CAD  = n : m. 

Through  A'B'  and  the  lines  of  division  of  Z C'A'D'  pass 
planes,  and  through  AB  and  the  lines  of  division  of 
Z CAD  pass  planes.  These  planes  will  divide  the  dihedral 
Z C'-A'B'-D'  into  n,  and  Z C-AB-D  into  m parts,  all 
equal.  Art.  552. 

/.  A C'-A'B'-D'  : Z C-AB-D  = n : m. 

Hence  Z C'-A'B’-D'  : Z C-AB-D  = A C'A'D'  : ACAD. 

(Why  T) 


DIHEDRAL  ANGLES 


341 


Case  II.  When  the  plane  angles  G'A'D'  and  CAD  (Figs. 
3 and  1)  are  incommensurable. 


Proof.  Divide  the  Z CAD  into  any  number  of  equal 
parts,  and  apply  one  of  these  parts  to  the  Z C'A'D’.  It 
will  be  contained  a certain  number  of  times  with  a remain- 
der, as  ALA'D',  less  than  the  unit  of  measure. 


Hence  the  A C' A' L and  CAD  are  commensurable. 


z C'—A'B'  — L\A  C—AB—D—  Z C'A'L:  A CAD.  Case  I- 
If  now  we  let  the  unit  of  measure  be  indefinitely  dimin- 
ished, the  ALA'D’,  which  is  less  than  the  unit  of  measure, 
will  be  indefinitely  diminished. 

.-.  Z C'A'L  = Z C’A'D’  as  a limit,  and 

Z C —A' B’  — L A Z C ' — A' B'  — D'  as  a limit.  Art.  251. 


Hence 


ZC'-A'B'-L 
Z C-AB-D 


Z C'-A'B'-D ' 


Z C-AB-D 

Z C'A'L 


as  its  limit; 


becomes  a variable,  with 


Art.  253,  3. 


Also 


ACAD 


limit. 


Z C'A'D' 

becomes  a variable  with  — . as  its 

Z CAD 

Art.  253,  3. 

Z C'A'L 


But  the  variable  — vr  = the  variable  , _ . _ 

AC-AB-D  ACAD 

always.  Case  I. 

„ A C'-A'B'-D'  n ,.  A C'A'D' 

the  limit  — — — . T,  ^ =the  limit  — _ • Art.  254. 


Z C-AB-D 


ACAD 


Q.  E.  D. 


Ex.  1.  How  many  straight  lines  are  necessary  to  indicate  a dihe- 
dral angle  (as  ZE-AB-D,  p.  338)?  How  many  straight  lines  are 
necessary  to  indicate  the  plane  angle  of  a dihedral  angle  ? Hence, 
what  is  the  advantage  of  using  a plane  angle  of  a dihedral  angle 
instead  of  the  dihedral  angle  itself  t 

Ex.  2.  Give  three  additional  properties  of  dihedral  angles  anal- 
ogous to  properties  of  plane  angles  given  in  Book  I. 


342 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XVII.  Theorem 

555.  If  a straight  line  is  perpendicular  to  a plane,  every 
plane  drawn  through  that  line  is  perpendicular  to  the  plane. 

P 


Given  the  line  AB  J_  plane  MN,  and  the  plane  PQ 
passing  through  AB  and  intersecting  HOT  in  RQ. 

To  prove  PQ  _L  MN. 

Proof.  In  the  plane  MN  draw  BC  _L  RQ  at  B. 

But  AB  J_  RQ.  Art.  505. 

.*.  Z ABC  is  the  plane  Z of  the  dihedral  ZP-RQ-M. 

Art.  548. 

But  ZABC  is  a right  Z , Art.  505. 

{for  AB  J_  MN  by  hyp.). 

PQ  -L  MN.  Art.  547. 

Q.  E.  D. 


556.  Cor.  A plane  perpendicular  to  the  edge  of  a 
dihedral  angle  is  perpendicular  to  each  of  the  two  faces  form- 
ing the  dihedral  an  ale, 


DIHEDRAL  ANGLES 


343 


Proposition  XVIII.  Theorem 

557.  If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to  their 
line  of  intersection  is  perpendicular  to  the  other  plane. 


Given  the  plane  PQ  X plane  BIN  and  intersecting  it  in 
the  line  BQ;  and  AB  a line  in  PQ  _L  BQ. 

To  prove  AB  X plane  MB. 

Proof.  In  the  plane  MN  draw  BC  _L  BQ. 

:.  /.ABC  is  the  plane  Z of  the  dihedral  /P-BQ-M. 

Art.  548. 

.-.  Z ABC  is  a rt.  Z , Art.  554. 

( for  P-BQ-M  is  a right  dihedral  Z ). 

.'.  AB  _L  BC  and  BQ  at  their  intersection. 

AB  X plane  MN.  (Why  ?) 

Q.  E.  D. 

558.  Cor.  1.  If  two  planes  are  perpendicular  to  each 
other,  a perpendicular  to  one  of  them  at  any  point  of  their 
intersection  will  lie  in  the  other  plane. 

For,  in  the  above  figure,  a X erected  at  the  point  B in 
the  plane  MN  must  coincide  with  AB  lying  in  the  plane 
PQ  and  X MN,  for  at  a given  point  in  a plane  only  one  X 
can  be  drawn  to  that  plane  (Art.  515). 

559.  Ccr.  2.  If  two  planes  are  perpendicular  to  each 
other,  a perpendicular  to  one  plane,  from  a point  in  the  other 
plane,  will  lie  in  the  other  plane, 


344  BOOK  VI.  SOLID  GEOMETRY 

Proposition  XIX.  Theorem 

560.  If  two  intersecting  planes  are  each  perpendicular 
to  a third  plane.,  their  line  of  intersection  is  perpendicular 
to  the  third  plane. 


Given  the  planes  PQ  and  BS  _L  plane  MN,  and  inter- 
secting in  the  line  AB. 

To  prove  AB  _L  plane  MN. 

Proof.  At  the  point  B in  which  the  three  planes  meet 
erect  a _L  to  the  plane  MN.  This  J_  must  lie  in  the  plane  PQ, 
and  also  in  the  plane  BS.  Art  55S- 

Hence  this  _L  must  coincide  with  AB,  the  intersection 
of  PQ  and  BS.  Art.  508,  2. 

.'.  AB  -L  plane  MN. 

Q.  E.  D. 

561.  Cor.  If  two  planes,  including  a right  dihedral 
angle,  are  each  perpendicular  to  a third  plane,  the  intersec- 
tion of  any  two  of  the  planes  is  perpendicular  to  the  third 
plane,  and  each  of  the  three  lines  of  intersection  is  perpen- 
dicular to  the  other  two. 


Ex.  1.  Name  all  the  dihedral  angles  on  the  above  figure. 

Ex.  2.  If  Z CBQ=30°,  find  the  ratio  of  each  pair  of  dihedral  Z. 


DIHEDRAL  ANGLES 


345 


Proposition  XX.  Theorem 


Given  plane  CB  bisecting  the  dihedral  AA-BR-P,  P 
any  point  in  plane  BC,  PQ  and  PT  _L  faces  BA  and  BP, 
respectively. 

To  prove  PQ=PT. 

Proof.  Through  PQ  and  PT  pass  a plane  intersecting 


AB  in  QR,  BP  in  ET.and  BG  in  PR. 

Then  plane  PQT  _L  planes  AB  and  BP.  Art.  555. 

plane  PQT  _L  line  RB,  the  intersection  of  the  planes 
AB  and  BP.  Art.  560. 

RB  JL  RQ,  RP  and  RT.  Art.  505. 

A QRP  and  PRT  are  the  plane  A of  the  dihedral 
A A-BR-P  and  P-BR-P.  Art.  548. 

Bnt  these  dihedral  A are  equal.  Hyp. 

.*.  Z QRP—  A PRT.  Art.  554. 

.*.  rt.A  PQR  = rt.  A PRT.  (Why?) 

.*.  PQ=PT.  (Why?) 


Q.  E.  D. 

563.  The  locus  of  all  points  equidistant  from  the  faces 
of  a dihedral  angle  is  the  plane  bisecting  the  dihedral  angle. 


562.  Every  point  in  the  plane  which  bisects  a given 
dihedral  angle  is  equidistant  from  the  faces  of  the  dihedral 

n.nnl.p. 


346 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XXI.  Problem 

564.  Through  any  straight  line  not  perpendicular  to  a 
given  plane , to  pass  a plane  perpendicular  to  the  given  plane. 


Given  the  line  AB  not  J-  plane  MN. 

To  construct  a plane  passing  through  AB  and  _L  MN. 

Construction.  From  a point  A in  the  line  AB  draw  a _L 
AC  to  the  plane  MN.  Art.  516. 

Through  the  intersecting  lines  AB  and  AC  pass  the 
plane  AD.  Art.  503. 

Then  AD  is  the  plane  required. 

Proof.  The  plane  AD  passes  through  AB.  Constr. 

Also  plane  AD  _L  plane  MN,  Art.  555. 

( for  it  contains  AC,  which  is  _L  MN). 

Q.  E.  F. 

565.  Cor.  1.  Through  a straight  line  not  perpendicular 
to  a given  plane  only  one  plane  can  he  passed  perpendicu- 
lar to  that  plane. 

For,  if  two  planes  could  be  passed  through  AB  _L  plane 
MN,  this  intersection  AB  would  be  _L  MN  (Art.  560), 
which  is  contrary  to  the  hypothesis. 

566.  Cor.  2.  The  projection  upon  a plane  of  a straight 
line  not  perpendicular  to  that  plane  is  a straight  line. 

For,  if  a plane  be  passed  through  the  given  line  _L  to 
the  given  plane,  the  foot  of  a 1 from  any  point  in  the 
line  to  the  given  plane  will  be  in  the  intersection  of  the 
two  planes  (Art.  559), 


DIHEDRAL  ANGLES 


347 


Proposition  XXII.  Theorem 

567.  flit  acute  angle  which  a line  makes  ivith  its  pro- 
jection on  a plane  is  the  least  angle  which  it  makes  ivith  any 
line  of  the  plane  through  its  foot. 


Given  line  AB  meeting  the  plane  MN  in  the  point  B, 
BG  the  projection  of  AB  on  MN,  and  PB  any  other  line 
in  the  plane  MN  through  B. 

To  prove  that  Z ABC  is  less  than  Z ABP. 

Proof.  Lay  off  PB  equal  to  CB,  and  draw  AG  and  AP. 


Then,  in  the  A ABC  and  ABP, 

AB  — AB.  (Why?) 

BC=BP.  (Why?) 

But  AC  < AP.  Art.  521. 

Z ABC  is  less  than  Z ABP  Art.  108. 


Q.  E.  D. 

568.  Def.  The  inclination  of  a line  to  a plane  is  the 

acute  angle  which  the  given  line  makes  with  its  projection 
upon  the  given  plane. 


Ex.  1.  A plane  has  an  inclination  of  47°  to  each  of  the  faces  of  a 
dihedral  angle  and  is  parallel  to  the  edge  of  the  dihedral  angle;  how 
many  degrees  are  in  the  plane  angle  of  the  dihedral  angle? 

Ex.  2.  In  the  figure  on  page  345,  if  PT=QT,  how  large  is  the 
dihedral  L A~BR-D“l  if  PT—llT.  how  large  is  it? 


348 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XXIII.  Problem 

569.  To  draw  a common  perpendicular  to  any  two  lines 
not  in  the  same  plane. 


Given  the  lines  AB  and  CD  not  in  the  same  plane.  ‘ 

To  construct  a line  perpendicular  to  both  AB  and  CD. 
Construction.  Through  AB  pass  a plane  MX ||  line  CD. 

Art.  529. 

Through  CD  pass  a plane  CF  _L  plane  d/X^Art.  564),  and 
intersecting  plane  MX  in  the  line  EF. 

Then  EF  \\  CD  (Art.  528),  .*.  EF  must  intersect  AB 
(which  is  not  ||  CD  by  hyp.)  in  some  point  K. 


At  E in  the  plane  CF  draw  LK  _L  EF.  Art.  »74. 

Then  BE  is  the  perpendicular  required. 

Proof.  EE  J_  EF.  Constr. 

/.  LE  _L  CD.  Art.  123. 

Also  LE  J_  plane  MX.  Art.  557. 

.*.  LE  _L  line  AB.  (Why?) 


/.  LE  JL  both  CD  and  AB. 

y.  E.  F. 

570.  Only  one  perpendicular  can  be  drawn  between  two 
lines  not  in  the  same  plane. 

For,  if  possible,  in  the  above  figure  let'another  line  BD 
be  drawn  _L  AB  and  CD.  Then,  if  a line  be  drawn  through 
B ||  CD,  BD  _L  this  line  (Art.  123),  and  .*.  _L  plane  MN 
(Art.  509).  Draw  DF  _L  line  EF-,  then  DF  ± plane  MX 
(Art.  557) . Hence  from  the  point  D two  A , DB  and  DF, 
are  drawn  to  the  plane  MX,  which  is  impossible  (Art.  517). 


POLYHEDRAL  ANGLES 


349 


POLYHEDRAL  ANGLES 

571.  A polyhedral  angle  is  the  amount 
of  opening  between  three  or  more  planes 
meeting  at  a point. 

Such  an  angle  may  be  regarded  as  a portion  of 
space  cut  out  by  the  planes  forming  the  angle. 


572.  The  vertex  of  a polyhedral  angle 
is  the  point  in  which  the  planes  forming  the  angle  meet;  the 
edges  are  the  lines  in  which  the  planes  intersect;  the  faces 
are  the  portions  of  the  planes  forming  the  polyhedral  angle 
which  are  included  between  the  edges;  the  face  angles  are 
the  angles  formed  by  the  edges. 

Each  two  adjacent  faces  of  a polyhedral  angle  form  a 
dihedral  angle. 

The  parts  of  a polyhedral  angle  are  its  face  angles  and 
dihedral  angles  taken  together. 


573.  Naming  a polyhedral  angle.  A polyhedral  angle 
is  named  either  by  naming  the  vertex,  as  V ; or  by  naming 
the  vertex  and  a point  on  each  edge,  as  V-ABC. 

In  case  two  or  more  polyhedral  angles  have  the  same 
vertex,  the  latter  method  is  necessary. 

In  the  above  polyhedral  angle,  the  vertex  is  V ; the 
edges  are  YA,  YB,  VC-,  the  face  angles 
are  AYB,  BYC,  AVC. 

574.  A convex  polyhedral  angle  is  a 

polyhedral  angle  in  which  a section 
made  by  a plane  cutting  all  the  edges 
is  a convex  polygon,  as  Y- ABODE. 

bib.  A trihedral  angle  is  a polyhe- 
dral angle  having  three  faces;  a tetrahedral  angle  is  one 
having  four  faces,  etc. 


350 


BOOK  VB,  SOLID  GEOMETSY 


576.  A trihedral  angle  is  rectangular,  birectangular,  or 
trirectangular,  according  as  it  contains  one,  two,  or  three 
right  dihedral  angles. 


577.  An  isosceles  trihedral  angle  is  a trihedral  angle 
two  of  whose  face  angles  are  equal. 


578.  Vertical  polyhedral  angles  are  polyhedral  angles 
having  the  same  vertex  and  the  faces  of  one  the  faces  of 
the  other  produced. 


579.  Two  equal  polyhe- 
dral angles  are  polyhedral  an- 
gles having  their  correspond- 
ing parts  equal  and  arranged 
in  the  same  order,  as  Y-ABC 
and  V'-A'B'C'. 

Two  equal  polyhedral  angles 
maybe  made  to  coincide. 

580.  Two  symmetrical  poly- 
hedral angles  are  polyhedral  an- 
gles having  their  corresponding 
parts  equal  but  arranged  in 


reverse  order. 


If  the  faces  of  a trihedral  angle,  Y-ABC,  be 
produced,  they  will  form  a vertical  trihedral  angle, 
Y-A'B'C,  which  is  symmetrical  to  V-ABC.  For, 
if  Y-A'B'C  be  rotated  forward  about  a horizontal 
axis  through  V,  the  two  trihedral  angles  are  seen 
to  have  their  corresponding  parts  equal  but  ar- 
ranged in  reverse  order. 

Similarly,  any  two  vertical  polyhedral  angles 
are  symmetrical. 


POLYHEDKAL  ANGLES 


351 


581.  Equivalence  of  symmetrical  polyhedral  angles.  It 
has  been  shown  in  Plane  Geometry  (Art.  488)  that  two 
triangles  (or  polygons)  symmetrical  with  respect  to  an 
axis  have  tlieir  corresponding  parts  equal  and  ax-ranged 
in  reverse  order.  By  sliding 


they  cannot  be  made  to  coincide, 

J ' a a 

but  by  lifting  one  of  them  up 

from  the  plane  in  which  it  lies  and  turning  it  over  it  may 
be  made  to  coincide  with  the  other  figure. 

Symmetrical  polyhedral  angles,  however,  cannot  be 
made  to  coincide  in  any  way;  hence  some  indirect  method 
of  showing  their  equivalence  is  necessary.  See  Ex.  29, 
p.  358,  and  Arts.  789-792. 


Ex.  1.  Name  the  trihedral  angles  on  the  figure  to  Prop.  XX.  If 
Z.PRQ= 90°,  what  kind  of  trihedral  angles  are  those  on  the  figure  ? 
If  /.PRQ^ 30°,  what  kind  are  they? 

Ex.  2.  Are  two  trireetangular  trihedral  angles  necessarily  equal  ? 
Prove  this. 

Ex.  3.  Are  two  lines  which  are  perpendicular  to  the  same  plane 
necessarily  parallel  ? Are  two  planes  which  are  perpendicular  to  the 
same  plane  necessarily  parallel  ? Are  two  planes  which  are  perpen- 
dicular to  the  same  line  necessarily  parallel  1 

Ex.  4.  Let  the  pupil  cut  out  three  pieces  of  pasteboard  of  the  form 
indicated  in  the  accompanying  figures ; cut  them  half  through  where 
the  lines  are  dotted;  fold  them  and  fasten  the  edges  so  as  to  form 
three  trihedral  angles,  two  of  which  (Figs.  1 and  2)  shall  be  equal 
and  two  (Figs.  1 and  3)  symmetrical.  By  experiment,  let  the  pupil 
find  which  pair  may  be  made  to  coincide,  and  which  not. 


two  such  figures  about  iu  a plane 


Pig-1 


Eg.3 


Eig,3 


352 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XXIV.  Theorem 

582.  The  sum  of  any  two  face  angles  of  a trihedral 
angle  is  greater  than  the  third  face  angle. 


S 


Given  the  trihedral  angle  S-ABC,  with  angle  ASC  its 
greatest  face  angle. 

To  prove  /ASB  + Z BSC  greater  than  /.ASC. 

Proof.  In  the  face  ASC  draw  SB,  making  /ASB  — 
/ ASB. 

Take  SB  = SB. 

In  the  face  ASC  draw  the  line  ABC  in  any  convenient 


direction,  and  draw  AB  and  BC. 

Then,  in  the  A ASB  and  ASB,  SA  — SA.  (Why?) 

SB  = SB,  and  ZASB=  /ASB.  (Why?) 

.*.  A ASB=A  ASB.  (Why?) 

.-.  AB=AB.  (Why?) 

Also  AB  + BC  > AC.  (Why?) 

Hence,  subtracting  the  equals  AB  and  AB. 

BC  > BC.  ( Why  T) 

Hence,  in  the  A BSC  and  BSC,  SC=SC,  SB  = SB,  and 
BC  > BC.  (Why  ?) 

.*.  /BSC  is  greater  than  /BSC.  Art.  108. 


To  each  of  these  unequals  add  the  equals  / ASB  and 
/ASB. 

:.  /ASB  + Z BSC  is  greater  than  / ASC.  (Why?) 

Q.  E.  D. 

Ex.  In  the  above  figure,  if  / ASC  equals  one  of  the  other  face 
angles  at  S,  as  Z ASB,  how  is  the  theorem  proved  ? 


POLYHEDRAL  ANGLES 


353 


Proposition  XXV.  Theorem 

583.  The  sum  of  the  face  angles  of  any  convex  polyhedral 
angle  is  less  than  four  right  angles. 


s 


Given  the  polyhedral  angle  S-ABCDE. 

To  prove  the  sum  of  the  face  A at  8 less  than  4 rt.  A . 
Proof.  Pass  a plane  cutting  the  edges  of  the  given  poly- 
hedral angle  in  the  points  A,  B,  C , B,  E. 

From  any  point  0 in  the  polygon  ABODE  draw  OA, 
OB,  OC,  OD,  OE. 

Denote  the  A having  the  common  vertex  S as  the  S A, 
and  those  having  the  common  vertex  0 as  the  0 A . 

Then  the  sum  of  the  A of  the  8 A=the  sum  of  A of 
the  0 A . Art.  134. 

But  Z SB  A + Z SBC  is  greater  than  Z ABC,  1 

Z SOB  + Z SOD  is  greater  than  Z BCD,  etc.  J 

.*.  the  sum  of  the  base  A of  the  S A > the  sum  of  the 
base  A of  the  0 A.  Ax.  9. 

. \ the  sum  of  the  vertex  A of  the  S A < the  sum  of 
the  vertex  A of  the  0 A,  Ax.  11. 

(if  unequals  be  subtracted  from  equals,  the  remainders  are  unequal 
in  reverse  order). 

But  the  sum  of  the  A at  0 = 4 rt.  A . (Why?) 

.'.  the  sum  of  face  A at  S < 4 rt.  A . Ax.  8. 

Q.  E.  0. 


W 


354 


BOOK  VI.  SOLID  GEOMETRY 


Proposition  XXVI.  Theorem 

584.  If  two  trihedral  angles  have  the  three  face  angles 
of  one  equal  to  the  three  face  angles  of  the  other,  the  tri- 
hedral angles  have  their  corresponding  dihedral  angles  equal, 
and  are  either  equal  or  symmetrical,  according  as  their' 
corresponding  face  angles  are  arranged  in  the  same  or  in 
reverse  order. 


Given  the  trihedral  A S-ABC  and  S’-A'B'C,  having 
the  face  A ASB,  ^.£(7  and  BSC  equal  to  the  face  A A'S'B', 
A'S'C'  and  B'S'C' , respectively. 

To  prove  that  the  corresponding  dihedral  A of  S-ABC 
and  S'-A'B'C'  are  equal,  and  that  A S-ABC  and  S'- 
A'B'C'  are  either  equal  or  symmetrical. 

Proof.  On  the  edges  of  the  trihedral  A take  SB, 
SC,  S' A',  S'B',  S' O'  all  equal. 

Draw  AB,  AC,  BC,  A'B',  A'C',  B'C'. 

Then,  1.  In  the  A ASB  and  A'S'B',  SA  = S'Ar,  SB  = 
S'B',  and  Z ASB=  A A'S'B'.  (Why?) 

.-.  A ASB=  A A'S'B'.  (Why?) 

/.  AB  = A'B'.  (Why?) 

2.  In  like  manner  AC=  A'C',  and  BC=B'C'. 

A ABC=A  A'B' V. 


(Why  ?) 


EXERCISES  ON  THE  LINE  AND  PLANE  355 


3.  Take  D a convenient  point  in  SA,  and  draw  DE  in 
the  face  A8B,  and  DF  in  the  face  ASC,  each  JL  SA. 

DE  and  DF  meet  AB  and  AC  in  points  E and  F, 
respectively,  (/or  £ SA£  and  gAC  are  aCMte)> 


Similarly,  take  S'D'^SD  and  construct  A D'E'F'. 

Then,  in  the  rt.  A ADE  and  A'D'E',  AD  — A'D and 
ADAE=D'A'E'.  (Why?) 

A ADE—  A A’D'E'.  (Why?) 

/.  AE=A’E',  and  DE=D'E’.  (Why?) 


4.  In  like  manner  it  maybe  shown  that  AF—A'F',  and 


DF=D'F. 

A AEF=  A A'E'F'.  (Why  ?) 

And  EF=E'F'  (Why?) 

5.  Hence,  in  the  A DEF  and  D'E'F',  DE—D'E',  DF 
= D'F'  and  EF=E'F'.  (Why?) 

/.  A DEF=  A D'E'F'.  (Why?) 

/.  ZEDF—  ZE'D'F'.  (Why?) 


But  these  A are  the  plane  A of  the  dihedral  A whose 
edges  are  SA  and  S' A'. 

dihedral  Z B-AS~C=  dihedral  AB'-A'S'-C  Art.  552. 

In  like  manner  it  may  be  shown  that  the  dihedral  A at 
SB  and  S'B'  are  equal;  and  that  those  at  SC  and  S'C'  are 
equal. 

the  trihedral  A S and  S'  are  either  equal  or  sym- 
metrical. Arts.  579,  580. 

Q.  E.  D. 

EXERCISES.  CROUP  64 

THEOREMS  CONCERNING  THE  LINE  AND  PLANE  IN  SPACE 

Ex.  1.  A segment  of  a line  not  parallel  to  a plane  is  longer  than 
its  projection  in  the  plane. 

Ex.  2.  Equal  straight  lines  drawn  from  a point  to  a plane  are 
equally  inclined  to  the  plane. 


356 


BOOK  VI.  SOLID  GEOMETRY 


Ex.  3.  A line  and  plane  perpendicular  to  the  same  plane  are 
parallel. 

Ex.  4.  If  three  planes  intersecting  in  three  straight  lines  are 
perpendicular  to  a plane,  their  lines  of  intersection  are  parallel. 


I 

\ 

\ 


Ex.  5.  If  a plane  bisects  any  line  at  right 
angles,  any  point  in  the  plane  is  equidistant 
from  the  ends  of  the  line. 

Ex.  6.  Given  AB  _L  plane  MN, 
and  AC  ± plane  RS; 
prove  BC  J.  NR. 


Ex.  7.  Given  PQ  ± plane  MN, 

PR  -L  plane  BL, 
and  RS  L plane  MN; 
prove  QS  i.  AB. 

Ex.  8.  If  a line  is  perpendicular  to  one 
of  two  intersecting  planes,  its  projection 
on  the  other  plane  is  perpendicular  to  the 
line  of  intersection  of  the  two  planes. 


Ex.  9.  Given  CE  J_  BE, 

AE  ± BE, 

and  Z C-AB-E  a rt.  dihedral  Z ; 
prove  CA  _L  plane  BAE. 

Ex.  1 0.  The  projections  of  two  parallel  lines  on 
a plane  are  parallel.  (Is  the  converse  of  this 
theorem  also  true  ?) 


Ex.  11.  If  two  parallel  planes  are  cut  by  two  non-parallel  planes, 
the  two  lines  of  intersection  in  each  of  the  parallel  planes  will  make 
equal  angles. 

Ex.  12.  If  a line  is  perpendicular  to  a plane,  any  plane  parallel 
to  the  line  is  perpendicular  to  the  plane.  (Is  the  converse  true  ?) 

Ex.  13.  In  the  figure  to  Prop.  VI,  given  AB  JL  MN  and  AF  J. 
BC;  prove  BE  -L  BC. 

Ex.  14.  Two  planes  parallel  to  a third  plane  are  parallel  to  each 

OtLfcl 

[Suo.  Draw  a line  X third  plane.) 


EXERCISES  ON  THE  LINE  AND  PLANE 


357 


Ex.  15.  The  projections  upon  a plane  of  two  equal  and  parallel 
straight  lines  are  equal  and  parallel. 

Ex.  16.  A line  parallel  to  two  planes  is  parallel  to  their  inter- 
section. 

Ex.  17.  In  the  figure  to  Prop.  XXII,  if  angle  CBP  is  obtuse, 
prove  lhe  angle  ABP  obtuse. 

Ex.  18.  In  a quadrilateral  in  space  (i.  e.,  a 
quadrilateral  whose  vertices  are  not  all  in  the 
same  plane),  show  that  the  lines  joining  the 
midpoints  of  the  sides  form  a parallelogram. 

Ex.  19.  The  lines  joining  the  midpoints  of 
the  opposite  sides  of  a quadrilateral  in  space  bisect  each  other. 

Ex.  20.  The  planes  bisecting  the  dihedral  angles  of  a trihedral 
angle  meet  in  a line  every  point  of  which  is  equidistant  from  the 
three  faces. 

[Sug.  See  Art.  562.] 

I Ex.  21.  Given  OQ  bisecting  /.BOS, 

PQ  _L  plane  EOS, 

QB  -L  OB, 
and  QS  L OS ; 
prove  PB=  PS,  PB  i_  OB, 
and  PS  -L  OS. 

Ex.  22.  In  a plane  bisecting  a given  plane  angle,  and  perpendicu- 
lar to  its  plane,  every  point  is  equidistant  from  the  sides  of  the  angle. 

[Sug.  See  Ex.  21;  or  through  P any  point  in  the  bisecting  plane 
pass  planes  L to  the  sides  of  the  Z , etc.] 

Ex.  23.  In  a trihedral  angle,  the  three  planes  bisecting  the  three 
face  angles  at  right  angles  to  their  respective  planes,  intersect  in  a line 
every  point  of  which  is  equidistant  from  the  three  edges  of  the  tri- 
hedral angle. 

Ex.  24.  If  two  face  angles  of  a trihedral  angle  are  equal,  the  dihe- 
dral angles  opposite  them  are  equal. 

Ex.  25.  In  the  figure  to  Prop.  XXIV,  prove  that  /ASC~\-  / BSC 
is  greater  than  /ASD  -j-  /BSD. 

Ex.  26.  The  common  perpendicular  to  two  lines  in  space  is  the 
shortest  line  between  them. 


358 


BOOK  YI.  SOLID  GEOMETEY 


Ex.  27.  Given  MN  ||  BS, 
and  PB  = pb ; 
prove  AABC=  Adbc, 
and  A ABC  =c=  A abc. 

Ex.  28.  Two  isosceles  symmetrical 
trihedral  angles  are  equal. 

Ex.  29.  Any  two  symmetrical  trihe- 
dral angles  are  equivalent. 

[Sue.  Take  SA,  SB,  SC,  S'A', 

S'B' , S'C1,  all  equal.  Pass  planes  ABC, 
A'B’C’.  Draw  SO  and  S'O'  ± these 
planes.  Then  the  trihedral  A are 
divided  into  three  pairs  of  isosceles  B 
symmetrical  trihedral  A,  etc.] 


EXERCISES.  CROUP  65 

LOCI  IN  SPACE 

Find  the  locus  of  a point  equidistant  from 

Ex.  1.  Two  parallel  planes.  Ex.  3.  Three  given  points. 

Ex.  2.  Two  given  points.  Ex.  4.  Two  intersecting  lines. 

Ex.  5.  The  three  faces  of  a trihedral  angle. 

Ex.  6.  The  three  edges  of  a trihedral  angle. 

Find  the  locus 

Ex.  7.  Of  all  lines  passing  through  a given  point  and  parallel  to 
a given  plane. 

Ex.  8.  Of  all  lines  perpendicular  to  a given  line  at  a given  ■point 
in  the  line. 

Ex.  9.  Of  all  points  in  a given  plane  equidistant  from  a given 
point  outside  the  plane. 

Ex.  10.  Of  all  points  equidistant  from  two  given  points  and  from 
two  parallel  planes. 

Ex.  11.  Of  all  points  equidistant  from  two  given  points  and  from 
two  intersecting  planes. 

Ex.  12.  Of  all  points  at  a given  distance  from  a given  plane  and 
equidistant  from  two  intersecting  lines. 


EXERCISES  ON  THE  LINE  AND  PLANE  359 


EXERCISES.  CROUP  66 

PROBLEMS  CONCERNING  THE  POINT,  LINE  AND  PLANS 
IN  SPACE 

Ex.  1.  Through  a given  point  pass  a plane  parallel  to  a given 
plane. 

Ex.  2.  Through  a given  point  pass  a plane  perpendicular  to  a 
given  plane. 

Ex.  3.  Through  a given  point  to  construct  a plane  parallel  to  two 
given  lines  which  are  not  in  the  same  plane. 

Prove  that  only  one  plane  can  be  constructed  fulfilling  the  given 
conditions. 

Ex.  4.  Bisect  a given  dihedral  angle. 

Ex.  5.  Draw  a plane  equally  inclined  to  three  lines  which  meet  at 
a point. 

Ex.  6.  Through  a given  point  draw  a line  parallel  to  two  given 
intersecting  planes. 

Ex.  7.  Find  a point  in  a plane  such  .that  lines  drawn  to  it  from 
two  given  points  without  the  plane  make  equal  angles  with  the  plane. 

[Sug.  See  Ex.  23,  p.  176.] 

Ex.  8.  Find  a point  in  a given  line  equidistant  from  two  given 
points. 

Ex.  9.  Find  a point  in  a plane  equidistant  from  three  given  points. 

Ex.  10.  Find  a point  equidistant  from  four  given  points  not  in  a 
plane. 

Ex.  11.  Through  a given  point  draw  a line  which  shall  intersect 
two  given  lines. 

[Sug.  Pass  a plane  through  the  given  point  and  one  of  the  given 
lines,  and  pass  another  plane  through  the  given  point  and  the  other 
given  line,  etc.] 

Ex.  12.  Through  a given  point  pass  a plane  cutting  the  edges  of 
a tetrahedral  angle  so  that  the  section  shall  be  a parallelogram. 

[Sug.  Produce  each  pair  of  opposite  faces  to  intersect  in  a 
straight  line,  etc.] 


Book  VII 

POLYHEDRONS 

585.  A polyhedron  is  a solid  bounded  by  planes. 


586.  The  faces  of  a polyhe- 

\irnm 

dron  are  its  bounding  planes;  the 

! w '—p & 

edges  of  a polyhedron  are  the  lines 

of  intersection  of  its  faces. 

1 mpM/k  ’ VBl 

! / ■pB  WW. 

A diagonal  of  a polyhedron  is 

IL Pfe  flF 

a straight  line  joining  two  of  its 

vertices  which  are  not  in  the 

same  face.  The  vertices  of  a poly-  polyhedron 

liedron  are  the  points  in  which  its 
edges  meet  or  intersect. 


587.  A convex  polyhedron  is  a polyhedron  in  which  a 
section  made  by  any  plane  is  a convex  polygon. 

Only  convex  polyhedrons  are  to  be  considered  in  this  book. 

588.  Classification  of  polyhedrons.  Polyhedrons  are 
sometimes  classified  according  to  the  number  of  their 
faces.  Thus,  a tetrahedron  is  a polyhedron  of  four  faces; 
a hexahedron  is  a polyhedron  of  six  faces;  an  octahedron 
is  one  of  eight,  a dodecahedron  one  of  twelve,  and  an 
icosahedron  one  of  twenty  faces. 


^Tetrahedron  Onbe  Octahedron  Dodecahedron  Icosahedron 

. (360) 


POLYHEDRONS 


361 


The  polyhedrons  most  important  in  practical  life  are  those  deter- 
mined by  their  stability,  the  facility  with  which  they  can  be  made  out 
of  common  materials,  as  wood  and  iron,  the  readiness  with  which 
they  can  be  packed  together,  etc.  Thus,  prism  means  "something 
lawed  off.” 


PRISMS  AND  PARALLELOPIPEDS 

589.  A prism  is  a polyhedron  bounded 
by  two  parallel  planes  and  a group  of 
planes  whose  lines  of  intersection  are 
parallel. 

590.  The  bases  of  a prism  are  the 
faces  formed  by  the  two  parallel  planes; 
the  lateral  faces  are  the  faces  formed  by 
the  group  of  planes  whose  lines  of  intersection  are  parallel. 

The  altitude  of  a prism  is  the  perpendicular  distance 
between  the  planes  of  its  bases. 

The  lateral  area  of  a prism  is  the  sum  of  the  areas  of  the 
lateral  faces. 

591.  Properties  of  a prism  inferred  immediately. 

1.  The  lateral  edges  of  a prism  are  equal,  for  they  are 
parallel  lines  included  between  parallel  planes  (Art.  589) 
and  are  therefore  equal  (Art.  532). 

2.  The  lateral  faces  of  a prism  are  parallelograms  (Art. 
160),  for  their  sides  formed  by  the  lateral  edges  are  equal 
and  parallel. 

3.  The  bases  of  a prism  are  equal  polygons,  for  their 
homologous  sides  are  equal  and  parallel,  each  to  each, 
(being  opposite  sides  of  a parallelogram),  and  their  homol- 
ogous angles  are  equal  (Art.  538). 

592.  A right  section  of  a prism  is  a section  made  by  a 
plane  perpendicular  to  the  lateral  edges. 


362 


BOOK  Vn.  SOLID  GEOMETRY 


593.  A triangular  prism  is  a prism  whose  base  is  a 
triangle  ; a quadrangular  prism  is  one  whose  base  is 
a quadrilateral,  etc. 


Oblique  Prisms  Eight  Prism  Eegular  Prism 

594.  An  oblique  prism  is  a prism  whose  lateral  edges 
are  oblique  to  the  bases. 


595.  A right  prism  is  a prism  whose 
lateral  edges  are  perpendicular  to  the  bases. 

596.  A regular  prism  is  a right  prism 
whose  bases  are  regular  polygons. 

597.  A truncated  prism  is  that  part  of 
a prism  included  between  a base  and  a 
section  made  by  a plane  oblique  to  the 
base  and  cutting  all  the  lateral  edges. 


Truncated  Prism 


598.  A parallelopiped  is  a prism  whose  bases  are  paral- 
lelograms. 

Hence,  all  the  faces  of  a parallelopiped  are  parallelograms. 


Oblique  Right  Rectangular  Cube 

Parallelopiped  Parallelopiped  Parallelopiped 


599.  A right  parallelopiped  is  a parallelopiped  whose 
lateral  edges  are  perpendicular  to  the  bases. 


PKISMS  AND  PAKALLELOPIPEDS 


363 


600.  A rectangular  parallelopiped  is  a right  parallelo- 
piped  whose  bases  are  rectangles. 

Hence,  all  the  faces  of  a rectangular  'parallelopiped  are 
rectangles. 

601.  A cube  is  a rectangular  parallelopiped  whose  edges 
are  all  equal. 

Hence,  all  the  faces  of  a exile  are  squares. 

602.  The  unit  of  volume  is  a cube  whose  edge  is  equal 
to  some  linear  unit,  as  a cubic  inch,  a cubic  foot,  etc. 

603.  The  volume  of  a solid  is  the  number  of  units  of 
volume  which  the  solid  contains. 

Being  a number , a volume  may  often  be  determined  from  other 
numbers  in  certain  expeditious  ways,  which  it  is  one  of  the  objects  of 
geometry  to  determine. 

604.  Equivalent  solids  are  solids  whose  volumes  are 
equal. 


Ex.  1.  What  is  the  least  number  of  faces  which  a polyhedron  can 
have  ? 

Ex.  2.  A square  right  prism  is  what  kind  of  a parallelopiped  ? 

Ex.  3.  Are  there  more  right  parallelopipeds  or  rectangular  paral- 
lelepipeds f That  is,  which  of  these  includes  the  other  as  a special 
case  ? 

Ex.  4.  Prove  that  if  a given  straight  line  is  perpendicular  to  a 
given  plane,  and  another  straight  line  is  perpendicular  to  another 
plane,  and  the  two  planes  are  parallel,  then  the  two  given  lines  are 
.parallel. 


364  BOOK  VII.  SOLID  GEOMETRY 

Proposition  I.  Theorem 

605.  Sections  of  a prism  made  by  parallel  planes  cutting 
all  the  lateral  edges  are  equal  polygons. 


Given  tlie  prism  PQ  cut  by  ||  planes  forming  the  sections 
AD  and  A'D'. 


To  prove  section  AD  = section  A'D'. 

Proof.  AB,  BC,  CD,  etc.,  are  [|  A'B' , B'C',  CD',  etc., 
respectively.  Art.  531. 

.'.  AB,  BC,  CD,  etc.,  are  equal  to  A'B',  B'C',  CD', 
etc.,  respectively.  Art- 157. 


Also  A ABC,  BCD,  etc.,  are  equal  to  A A' B'C',  B'CD', 
etc.,  respectively.  Art. 538 

.-.  A B CDE — A'B' CD' E' , Art,  47. 

( for  the  polygons  have  all  their  parts  equal,  each  to  each,  and  .’.  can  he 
made  to  coincide). 

Q.  E.  D. 


606.  Cor.  1.  Every  section  of  a prism  made  by  a plane 
parallel  to  the  base  is  equal  to  the  base. 


607.  Cor.  2.  All  right  sections  of  a prism  are  equal. 


PRISMS 


365 


Proposition  II,  Theorem 

608.  The  lateral  area  of  a prism  is  equal  to  the  product 
of  the  perimeter  of  a right  section  by  a lateral  edge. 


Given  the  prism  BQ,  with  its  lateral  area  denoted  by  8 
and  lateral  edge  by  E;  and  AD  a right  section  of  the  given 
prism  with  its  perimeter  denoted  by  P. 

To  prove  8—PX  E. 

Proof.  In  the  prism  BQ,  each  lateral  edge  = P.  Art.  591, 1. 

Also  AB  A GPL,  BG  A IJ,  etc.  Art.  505. 

Hence  area  ZZ7  BE=AB  X GD=  AB  X _E','| 

areaZA7  GJ=BGX  E,  r Art.  385. 

area dJ  IQ  = CDXE,  etc.  J 

But  8,  the  lateral  area  of  the  prism,  equals  the  sum  of 
the  areas  of  the  AE7  forming  the  lateral  surface. 

/.  adding,  S=  (AB  + BG  + CD  + etc. ) X E.  Ax.  2. 

Or  S=P  X E. 

Q.  E.  D. 

609.  Cor.  The  lateral  area  of  a right  prism  equals  the 
product  of  the  perimeter  of  the  base  by  the  altitude. 


Ex.  Find  the  lateral  area  of  a right  prism  whose  altitude  is  12  in., 
and  whose  base  is  an  equilateral  triangle  with  a side  of  6 in.  Also 
find  the  total  area  of  this  figure. 


3C6 


BOOK  VII.  SOLID  GEOMETKY 


Proposition  III.  Theorem 

610.  If  two  prisms  have  the  three  faces  including  a 
trihedral  angle  of  one  equal , respectively , to  the  three  faces 
including  a trihedral  angle  of  the  other,  and  similarly 
placed,  the  prisms  are  equal. 


Given  the  prisms  AJ  and  A'J' , having  the  faces  AK, 
AD,  AG  equal  to  the  faces  A'K' , A'D' , A'G' , respectively, 
and  similarly  placed. 

To  prove  AJ—A'J'. 

Proof.  The  face  A EAF,  FAB  and  BAF  are  equal, 
respectively,  to  the  face  A E'A'F',  E'A'B ' and  B'A'F'.  Hyp. 

trihedral  AA  = trihedral  A A'  Art.  584. 

Apply  the  prism  A'J'  to  the  prism  AJ,  making  each  of 
the  faces  of  the  trihedral  A A'  coincide  with  corresponding 
equal  face  of  the  trihedral  A A.  Geom.Ax.  2. 

A the  plane  F'J'  will  coincide  in  position  with  the 

plane  FJ , Art. 500. 

(for  the  points  O’ , F , E'  coincide  with  G,  F,  E,  respectively) . 

Also  the  point  C will  coincide  with  the  point  C. 

C'iPwill  take  the  direction  of  CH.  Geom.Ax.  3. 
.’.  E'  will  coincide  with  R.  Art.  508,  1. 

In  like  manner  J'  will  coincide  with  J. 

Hence  the  prisms  AJ  and  A'J'  coincide  in  all  points. 

.'.  AJ  = A'J'.  Art.  47. 

611.  Cor.  1.  Two  truncated  prisms  are  equal  if  the 
three  faces  including  a trihedral  angle  of  one  are  equal  to 
the  three  faces  including  a trihedral  angle  of  the  other. 


PKISIMS 


367 


612.  Cor.  2.  Two  right  prisms  are  equal  if  they  have 
equal  bases  and  equal  altitudes.  ^ 

Proposition  IV.  Theorem  yA 

613.  An  oblique  prism  is  equivalent  to  a right  prism 
whose  base  is  a right  section  of  the  oblique  prism  and  whose 
altitude  is  equal  to  a lateral  edge  of  the  oblique  prism. 


Given  the  oblique  prism  AD ',  with  the  right  section  FJ; 
also  the  right  prism  FJ'  whose  lateral  edges  are  each  equal 
to  a lateral  edge  of  AD' . 

To  prove  AD'  -- c=  FJ' . 

Proof.  AA'  = FF' . Hyp. 

Subtracting  FA'  from  each  of  these,  AF—A'F'.  (Why?) 
Similarly  BG  = B'G'. 

Also  AB  = A'B',  and  FG—F'G'.  Art.  155. 

And  A of  face  AG  = homologous  A of  fac eA'G'.  Art.  iso. 

face  AG  — face  A'G',  Art.  47  i 

( for  they  have  all  their  parts  equal,  each  to  each,  and  can  he  made 
to  coincide). 

In  like  manner  face  AIT=  face  A'K' . 

But  face  AD— face  A’D' . Art.  591,  3. 

.*.  truncated  prism  AJ=  truncated  prism  A'J'.  Art.  611. 
To  each  of  these  equals  add  the  solid  FD' . 

AD'^FJ'.  (Why  ?) 

Q.  £.  D. 


368 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  V.  Theorem 

614.  The  opposite  lateral  faces  of  a parallelopiped  are 
equal  and  parallel. 


Given  the  parallelopiped  AH  with  the  base  AC. 
To  prove  AG  = and  ||  EH,  and  A J=  and  ||  BH. 


Proof.  The  base  AC  is  a HO  . Art.  598. 

.’.  AB  = and  ||  EC.  (Why?) 

Also  the  lateral  face  AJ  is  a ED  . Art.  591,  2. 

.\  AjP=and  ||  EJ.  (Why?) 

A IBAF=  l CEJ.  Art.  538. 


And  EDAG=EDEH.  Art.  162. 

Also  plane  AG'  1 1 plane  EH.  Art.  538. 

In  like  manner  it  may  be  shown  that  AJ  and  BH  are 
equal  and  parallel. 

Q.  E.  D. 


616.  Cor.  Any  two  opposite  faces  of  a parallelopiped 
may  he  taken  as  the  bases. 


Ex.  1.  How  many  edges  has  a parallelopiped  ? How  many  faces  ? 
How  many  dihedral  angles  ? How  many  trihedral  angles  ? 

Ex.  2.  Find  the  lateral  area  of  a prism  whose  lateral  edge  is  10 
and  whose  right  section  is  a’triangle  whose  sides  are  6,  7,  8 in. 

Ex.  3.  Find  the  lateral  area  of  a right  prism  whose  lateral  edge  is 
16  and  whose  base  is  a rhombus  with  diagonals  of  6 and  8 in. 


PRISMS 

Proposition  VI.  Theorem 


369 


616.  A plane  passed  through  two  diagonally  opposite 
ges  of  a parallelopiped  divides  the  parallelopiped  into  tiuo 
ivalent  triangular  prisms. 


Given  the  parallelopiped  AH  with  a plane  passed  through 
diagonally  opposite  edges  AF  and  GH,  forming  the  tri- 
gular  prisms  ABC-G  and  ADC-K. 

To  prove  ABC-G-ADC-K. 

Proof.  Construct  a plane  _L  to  one  of  the  edges  of  the 
ism  forming  the  right  section  PQBS,  having  the  diagonal 
formed  by  the  intersection  of  the  plane  FHCA. 

Then  PQ  ||  SR,  and  QR  ||  PS.  (Why?) 

PQRS  is  a OH  . (Why  ?) 

A PQR=  A PSR.  (Why  ?) 

But  the  triangular  prism  ABC-G  =c=  a prism  whose  base 
the  right  section  PQR  and  whose  altitude  is  AF.  Art.  613. 

Also  the  triangular  prism  ADC-K  =c=  a prism  whose  base 
s the  right  section  PSR  and  whose  altitude  is  AF.  (Why  ?) 

But  the  prisms  having  the  equal  bases,  PQR  and  PSR, 
d the  same  altitude,  AF,  are  equal.  Art.  612. 

/.  ABC-G  ADC-K.  Ax  l. 

Q.  £•  D. 


X 


370 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  VII.  Theorem 


617.  If  tivo  rectangular  parallelepipeds  have  equal  bases, 
they  are  to  each  other  as  their  altitudes. 


m 


Given  the  rectangular  parallelopipeds  P'  and  P havgj 
equal  bases  and  the  altitudes  A'B'  and  AB. 

To  prove  P'  : P=A'B'  •.  AB. 

Case  I.  When  the  altitudes  A'B ' and  AB  are  c<M 
mensurable. 

Proof.  Find  a common  measure  of  A'B'  and  AB , 

AK,  and  let  it  be  contained  in  A'B'  n times  and  in  AB\ 
times. 

Then  A'B'  : AB  — n : m. 

Through  the  points  of  division  of  A'B'  and  AB  pal 
planes  parallel  to  the  bases. 

These  planes  will  divide  P'  into  n,  and  P into  m small 
rectangular  parallelopipeds,  all  equal.  Art.  612. 

.'.  P'  : P—n  : in. 

P'  : P=A'B>  : AB.  (Why  ?) 

Case  II.  When  the  altitudes  A'B'  and  AB  are  incom-u |M 
mensurable. 

Let  the  pupil  supply  the  proof,  using  the  method  of 
limits.  (See  Art.  554) . 


618.  Def.  The  dimensions  of  a rectangular  parallelo- 
piped  are  the  three  edges  which  meet  at  one  vertex. 


PARALLELOPIPEDS 


371 


619.  Cor.  If  two  rectangular  parallelopipeds  have  two 
dimensions  in  common , they  are  to  each  other  as  their  third 
dimensions. 


Proposition  VIII.  Theorem 

620.  Two  rectangular  parallelopipeds  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 


i.'  Given  the  rectangular  parallelopipeds  P and  P'  having 
* the  common  altitude  a,  and  the  dimensions  of  their  bases 
h,  c and  b',  c',  respectively. 


V 


To  prove 


P b X c 
P’  b'X  c'" 


Proof.  Construct  the  rectangular  parallelepiped,  Q,  whose 
altitude  is  a and  the  dimensions  of  whose  base  are  b and  d . 

Then  Art.  619- 

Q cf 


Also 

Multiplying 

equalities, 


Q_=b_ 

P'  bf‘ 

the  cori’esponding 

P b X c 
P'~  b'Xc'" 


(Why?) 

members  of  these 

Ax.  4. 

Q.  E.  D. 


621.  Cor.  Two  rectangular  parallelopipeds  having  one 
dimension  in  common  are  to  each  other  as  the  products  of  the 
other  two  dimensions. 


372 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  IX.  Theorem 

622.  Any  two  rectangular  parallelepipeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


P' 


Given  the  rectangular  parallelopipeds  P and  P'  having 
the  dimensions  a,  b,  c and  a',  b',  c',  respectively. 


To  prove 


P a X b X c 
P'~a'  X V X d 


Proof.  Consti-uct  the  rectangular  parallelopiped  Q hav- 
ing the  dimensions  a,  b,  d . 


Then 

Also 


P = c 
Q o' 

Q _ a X b 
P~a’  X V 


Art.  61B. 


Art.  621. 


Multiplying  the  corresponding  members  of  these 
equalities, 

P _ a X b X c 
P'~ a'  Xb'  X c'  - 

Q.  E.  D. 


Ex.  1.  Find  the  ratio  of  the  volumes  of  two  rectangular  parallelo- 
pipeds whose  edges  are  5,  G,  7 in.  and  7,  8,  9 in. 

Ex.  2.  Which  will  hold  more,  a bin  10x2x7  ft.,  or  one  8x 
4x5  ft.  ? 

Ex.  3.  How  many  bricks  8 x4  x 2 in.  are  necessary  to  build  a wall 
80  x 6 ft.  x 8 in,  f 


373 


PARALLELOPIPEDS 


Proposition  X.  Theorem 

623.  The  volume  of  a rectangular  parallelopiped  is  equal 
to  the  product  of  its  three  dimensions. 


P 


b 1 


Given  the  rectangular  parallelopiped  P having  the  three 
dimensions  a , b,  c. 


To  prove  volume  of  P=«X5Xc. 

Proof.  Take  as  the  unit  of  volume  the  cube  77,  whose 
edge  is  a linear  unit. 


Then 


P a X b X c 

U~1  X 1 X l' 


Art.  622. 


The  volume  of  P is  the  number  of  times  P contains  the 
P 

unit  of  volume  77,  or  — ■ Art.  603. 


.*.  volume  of  P=a  X b X c. 
For  significance  of  this  result,  see  Art.  2. 


Q.  E.  D. 


624.  Cor.  1.  The  volume  of  a cube  is  the  cube  of  its 
edge . 


625.  Cor.  2.  The  volume  of  a rectangular  parallelo- 
piped is  equal  to  the  product  of  its  base  by  its  altitude. 

Ex.  1.  Find  the  number  of  cubic  inches  in  the  volume  of  a cube 
whose  edge  is  1 ft.  3 in.  How  many  bushels  does  this  box  contain,  if 
1 bushel  = 2150.42  cu.  in.  ? 

Ex.  2.  The  measurement  of  the  volume  of  a cube  reduces  to  the 
measurement  of  the  length  of  what  single  straight  line  t 


374 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XI.  Theorem 

626.  The  volume  of  any  parallelopiped  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given  the  oblique  parallelopiped  P,  with  its  base  denoted 
by  B,  and  its  altitude  by  E. 

To  prove  volume  of  P—B  X E. 

Proof.  In  P produce  the  edge  CD  and  all  the  edges 
parallel  to  CD. 

On  CD  produced  take  FG=CD. 

Pass  planes  through  P and  G _L  the  produced  edges, 
forming  the  parallelopiped  Q,  with  the  rectangular  base 
denoted  by  B' . 

Similarly  produce  the  edge  G1  and  all  the  edges  ||  GI. 

Take  IK=  GI,  and  pass  planes  through  I and  K ± the 
edges  last  produced,  forming  the  rectangular  parallelopiped 
R,  with  its  base  denoted  by  B". 


Then 

P-QoR. 

Art.  613. 

Also 

B^B'  = B". 

Art.  3S6. 

But 

volume  of  R = B " X E. 

Art.  625. 

volume  of  P—B " X E. 

Ax.  1. 

Or 

volume  of  P—B  X E. 

Ax.  8. 

Q.  E.  D. 

[Outline  Proof.  P^Q^R=B"  XH^BXE.] 


prisms  375 

Proposition  XII.  Theorem 

627.  The  volume  of  a triangular  prism  is  equal  to  the 
product  of  its  base  by  its  altitude. 

0 

K 


Given  the  triangular  prism  PQR-M,  with  its  volume 
denoted  by  V,  area  of  base  by  B,  and  altitude  by  3. 

To  prove  Y—  B X 3. 

Proof.  Upon  the  edges  PQ,  QB,  QM,  construct  the 


parallelopiped  QK. 

Hence  $17=  twice  PQR-M.  Art.  616. 

But  volume  of  QK—  area  PQRT  X 3.  Art.  626. 

= 2 B X 3.  As.  8. 

twice  volume  PQR-M=2B  X H.  As.  l. 

volume  PQR-M=  B X 3.  As.  5. 

Q.  E.  D. 


Ex.  1.  If  the  altitude  of  a triangular  prism  is  18  in.,  and  the  base 
is  a right  triangle  whose  legs  are  6 and  8 in.,  find  the  volume. 

Ex.  2.  Find  the  volume  of  a triangular  prism  whose  altitude  is  24, 
and  the  edges  of  whose  base  are  7,  8,  9.  Also  find  the  total  surface. 


376 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XIII.  Theorem 


628.  The  volume  of  any  prism  is  equal  to  the  product  of 
its  base  by  its  altitude. 


L 


Given  the  prism  AK,  with  its  volume  denoted  by  V, 
area  of  base  by  B,  and  altitude  by  R. 

To  prove  Y—B  X R. 

Proof.  Through  any  lateral  edge,  as  AF,  and  the  diag- 
onals of  the  base,  AC  and  AD,  drawn  from  its  foot,  pass 
planes. 

These  planes  will  divide  the  prism  into  triangular 
prisms. 

Then  V,  the  volume  of  the  prism  AK,  equals  the  sum 


of  the  volumes  of  the  triangular  prisms.  Ax.  6. 

But  the  volume  of  each  triangular  prism  = its  base  X R. 
Hence  the  sum  of  the  volumes  of  the  A prisms  = the 
sum  of  the  bases  of  the  A prisms  X R. 

= B X R.  Ax.  8. 

Y=B  X R.  Ax.  l. 


Q.  E.  D. 

629.  Cor.  1.  Two  prisms  are  to  each  other  as  the  pro- 
ducts of  their  bases  by  their  altitudes;  prisms  having  equiva- 
lent bases  and  equal  altitudes  are  equivalent. 


630.  Cor.  2.  Prisms  having  equivalent  bases  are  to 
each  other  as  their  altitudes;  prisms  having  equal  altitudes 
are  to  each  other  as  their  bases . 


PYRAMIDS 


377 


PYRAMIDS 

631.  A pyramid  is  a polyhedron 
hounded  by  a group  of  planes  passing 
through  a common  point,  and  by  another 
plane  cutting  all  the  planes  of  the  group. 

632.  The  base  of  a pyramid  is  the 
face  formed  by  the  cutting  plane;  the 
lateral  faces  are  the  faces  formed  by  the 
group  of  planes  passing  through  a com- 
mon point;  the  vertex  is  the  common  point  through  which 
the  group  of  planes  passes;  the  lateral  edges  are  the  inter- 
sections of  the  lateral  faces. 

The  altitude  of  a pyramid  is  the  perpendicular  from  the 
vertex  to  the  plane  of  the  base. 

The  lateral  area  is  the  sum  of  the  areas  of  the  lateral 
faces. 

633.  Properties  of  pyramids  inferred  immediately. 

1.  The  lateral  faces  of  a pyramid  are  triangles  (Art. 
508,  2). 

2.  The  base  of  a pyramid  is  a polygon  (Art.  508,  2). 

634.  A triangular  pyramid  is  a pyramid  whose  base  is 
a triangle;  a quadrangular  pyramid  is  a pyramid  whose 
base  is  a quadrilateral,  etc. 

A triangular  pyramid  is  also  called  a tetrahedron,  for  it  has  four 
faces.  All  these  faces  are  triangles,  and  any  one  of  them  may  be 
taken  as  the  base. 

635.  A regular  pyramid 

is  a pyramid  whose  base  is  a 
regular  polygon,  and  the  foot 
of  whose  altitude  coincides 
with  the  center  of  the  base. 


378 


BOOK  VII.  SOLID  GEOMETRY 


636.  Properties  of  a regular  pyramid  inferred  immedi- 
ately. 

1.  The  lateral  edges  of  a regular  pyramid  are  equal , for 
they  are  oblique  lines  drawn  from  a point  to  a plane 
cutting  off  equal  distances  from  the  foot  of  the  per- 
pendicular from  the  point  to  the  plane  (Art.  518). 

2.  The  lateral  faces  of  a regular  pyramid  are  equal  isos- 
celes triangles. 

637.  The  slant  height  of  a regular  pyramid  is  the  alti- 
tude of  any  one  of  its  lateral  faces. 

The  axis  of  a regular  pyramid  is  its  altitude. 

638.  A truncated  pyramid  is  the  portion  of  a pyramid 
included  between  the  base  and  a section  cutting  all  the 
lateral  edges. 

639.  A frustum  of  a pyramid  is  the 

part  of  a pyramid  included  between  the 
base  and  a place  parallel  to  the  base. 

The  altitude  of  a frustum  of  a pyramid  is 
the  perpendicular  distance  between  the  planes  of  its  bases. 

640.  Properties  of  a frustum  of  a pyramid  inferred  im- 
mediately. 

1.  The  lateral  faces  of  a frustum  of  a pyramid  are 
trapezoids . 

2.  The  lateral  faces  of  a frustum  of  a regular  pyramid 
are  equal  isosceles  trapezoids. 

641.  The  slant  height  of  the  frustum  of  a regular  pyra- 
mid is  the  altitude  of  one  of  its  lateral  faces. 

Ex.  1.  Show  that  the  foot  of  the  altitude  of  a regular  pyramid 
coincides  with  the  center  of  the  circle  circumscribed  about  the  base. 

Ex.  2.  The  perimeter  of  the  midsection  of  the  frustum  of  a pyra- 
mid equals  one -half  the  sum  of  the  perimeters  of  the  bases, 


PYRAMIDS 


379 


Proposition  XIV.  Theorem 


642.  The  lateral  area  of  a regular  pyramid  is  equal  to 
half  the  product  of  the  slant  height  by  the  perimeter  of  the 
base. 


Given  0- ABCDF  a regular  pyramid  with  its  lateral 
area  denoted  by  S,  slant  height  by  L,  and  perimeter  of  its 
base  by  P. 

To  prove  S = J L X P. 

Proof.  The  lateral  faces  OAB,  OBG,  etc.,  are  equal 
isosceles  A.  Art.  636,  2. 

Hence  each  lateral  face  has  the  same  slant  height,  L. 
the  area  of  each  lateral  face  = £ L X its  base, 
the  sum  of  all  the  lateral  faces  = £ L X sum  of  bases. 
= iLXP. 

:.S=iLXP.  Ax.  8. 

Q.  E.  D. 

643.  Cor.  The  lateral  area  of 
the  frustum  of  a regular  pyramid  is 
equal  to  one-half  the  sum  of  the 
perimeters  of  its  bases  multiplied 
by  its  slant  height. 


Ex.  Find  the  lateral  area  of  a regular  square  pyramid  whose  slant 
height  is  32,  and  an  edge  of  whose  base  is  16.  Find  the  total 
area  also, 


380 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XV.  Theorem 
644.  If  a pyramid  is  cut  by  a plane  parallel  to  the  base , 

I.  The  lateral  edges  and  the  altitude  are  divided  pro - 
portionally ; 

II.  The  section  is  a polygon  similar  to  the  base. 


Given  the  pyramid  S-ABCDF,  with  the  altitude  SO  cut 
by  a plane  MU,  which  is  parallel  to  the  base  and  intersects 
the  lateral  edges  in  a,  b,  c,  d,  f and  the  altitude  in  o. 

m _ 8a  Sb  _ Sc  _ _ So 

To  prove  I.  aA  SB  sc  ”•  s0 ' 


II. 


SA 

The 


section 
ABCDF. 


abcdf  similar  to  the  base 


Proof.  I.  Pass  a plane  through  the  vertex  S |]  IfX. 
Then  SA,  SB,  SC ..  . SO,  are  lines  intersected  by  three 
||  planes. 

. = ®]l=—  = . . . Art.  539. 

SA  SB  SC  SO 

II.  ab  ||  AB.  (Why?) 

.*.  A Sab  and  SAB  are  similar.  Art.  328. 

In  like  manner  the  A Sbc , Scd,  etc.,  are  similar  to  the 
A SBC,  SCD,  etc.,  respectively. 

ab  = fSb_^  = ^ = 


AB  \ SB ) BC  \SCJ  CD 


etc. 


PYRAMIDS 


381 


That  is,  the  homologous  sides  of  abcdf  and  ABCDF  are 
proportional. 

Also  Zabc  — Z ABC,  Zbcd  = Z BCD,  etc.  Art.  538. 

section  abcdf  is  similar  to  the  base  ABCDF.  Art.  321. 

Q.  E.  D. 


645.  Cor.  1.  A section  of  a pyramid  parallel  to  the 
base  is  to  the  base  as  the  square  of  its  distance  from  the 
vertex  is  to  the  square  of  the  altitude  of  the  pyramid. 

' 9 

_ abcdf  ab 

For  — 

ABCDF  AB2 


(Why  ?) 


But 


ab  Sa  So  . ab  _ So 2 
AB  8A  SO  " AB 2 SO2 
abcdf  _ So 2 
" ABCDF~ so* 


(Why?) 
(Why  ?) 


646.  Cor.  2.  If  two  pyramids  having  equal  altitudes 
are  cut  by  a plane  parallel  to  their  bases  at  equal  distances 
from  the  vertices,  the  sections  have  the  same  ratio  as  the 
bases. 


Let  S~ ABCDF  and  V-PQB  be  two  pyramids  cut  as 
described. 

12 


Then 


abcdf  So “ pqr  Vt‘ 

; — = dzr  (Art.  645  : also  = ' (Why?) 

ABCDF  SO1  PQR  VI 2 


But  VT=SO,  and  Vt  = So. 

abcdf  pqr  abcdf  ABCDF 
*’*  ABCDF~  PQR  °r  pqr  ~ PQR 


(Why  ?) 
(Why?) 


647.  Cor.  3.  If  tw:  pyramids  have  equal  altitudes  and 
equivalent  bases,  sections  made  by  planes  parallel  to  the  bases 
at  equal  distances  from  the  vertices  are  equivalent. 


382 


BOOK  VII.  SO  jID  GEOMETRY 


Proposition  XVI.  Theorem 

648.  The  volume  of  a triangular  pyramid  is  the  limit  of 
the  sum  of  the  volumes  of  a series  of  inscribed,  or  of  a series 
of  circumscribed  prisms  of  equal  altitude , if  the  number  of 
prisms  be  indefinitely  increased. 


Given  the  triangular  prism  O-ABC  with  a series  of  in- 
scribed, and  also  a series  of  circumscribed  prisms,  formed  by 
passing  planes  which  divide  the  altitude  into  equal  parts,  and 
by  making  the  sections  so  formed  first  upper  bases,  then 
lower  bases,  of  prisms  limited  by  the  nest  parallel  plane. 

To  prove  O-ABG  the  limit  of  the  sum  of  each  series,  if 
the  number  of  prisms  in  each  be  indefinitely  increased. 

Proof.  Each  inscribed  prism  equals  the  circumscribed 
prism  immediately  above  it.  Art.  629. 

/.  (sum  of  circumscribed  prisms)  — (sum  of  inscribed 
prisms)  = lowest  circumscribed  prism,  or  ABC-K. 

If  the  number  of  prisms  be  indefinitely  increased,  the 
altitude  of  each  approaches  zero  as  a limit. 

Hence  volume  ABC~K= 0,  Art.  253,  2. 

( for  its  base,  ABC,  is  constant  while  its  altitude  = 0). 

/.  (sum  of  circumscribed  prisms) — (sum  of  inscribed 
prisms)  a 0. 

.*.  volume  O-ABC — (either  series  of  prisms)  a 0. 

( for  this  difference  < difference  between  the  two  series,  which 
last  difference  0 ) . 

/.  O-ABC  is  the  limit  of  the  sum  of  the  volumes  of 
either  series  of  prisms.  q.  e.  d. 


PYRAMIDS 


383 


Proposition  XVII.  Theorem 

649.  If  two  triangular  pyramids  have  equal  altitudes 
and  equivalent  hases , they  are  equivalent. 

o o' 


Given  the  triangular  pyramids  O-ABG  and  O'-A'B'G' 
having  equivalent  bases  ABG  and  A'B'C',  and  equal 
altitudes. 

To  prove  O-ABG-O'-A'B'C . 

Proof.  Place  the  pyramids  so  that  they  have  the  com- 
mon altitude  R,  and  divide  E into  any  convenient  num- 
ber of  equal  parts. 

Through  the  points  of  division  and  parallel  to  the  plane  of 
the  bases  of  the  pyramids,  pass  planes  cutting  the  pyramids. 

Using  the  sections  so  formed  as  upper  bases,  inscribe  a 
series  of  prisms  in  each  pyramid,  and  denote  the  volumes 
of  the  two  series  of  prisms  by  V and  V . 

The  sections  formed  by  each  plane,  as  KLM  and  K'L'M 
are  equivalent.  r Art.  647. 

.*.  each  prism  in  O-ABG  =o=  corresponding  prism  in 
O’-A’B'G ' (Art.  629).  .'.  V=  V'.  Ax.  2. 

Let  the  number  of  parts  into  which  the  altitude  is 
divided  be  increased  indefinitely. 

Then  V and  V become  variables  with  O-ABG  and 
O'-A'B'G'  as  their  respective  limits.  Art.  648. 

But  F=o  V'  always.  (Why?) 

O-ABG ~ O’-A'B'G’.  (Why  ?) 

Q.  E.  D. 


384 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XVIII.  Theorem 

650.  The  volume  of  a triangular  pyramid  is  equal  to 
one-tliird  the  product  of  its  base  by  its  altitude. 


Given  the  triangular  pyramid  O-ABC,  having  its  volume 
denoted  by  V,  the  area  of  its  base  by  B,  and  its  altitude  by  E. 

To  prove  Y=  $ B X H. 

Proof.  On  ABC  as  a base,  with  OB  as  a lateral  edge, 
construct  the  prism  ABC-BOF. 

Then  this  prism  will  be  composed  of  the  original  pyra- 
mid O-ABC  and  the  quadrangular  pyramid  O-ADFC. 

Through  the  edges  OB  and  OC  pass  a plane  intersecting 
the  face  ABFC  in  the  line  BC,  and  dividing  the  quadrangular 
pyramid  into  the  triangular  pyramids  O-ABC  and  O-BFC. 

Then  O-ABC-O-BFC,  Art.  649. 

(for  they  have  the  common  vertex  O,  and  the  equal  bases  ADC  and  DFC). 

But  O-BFC  may  be  regarded  as  having  C as  its  vertex 
and  BOF  as  its  base.  Art.  634. 

O-BFC— O-ABC.  Art.  649.. 

.’.  the  prism  is  made  up  of  three  equivalent  pyramids. 

.'.  O-ABC  = J the  prism.  Ax.  5. 

But  volume  of  prism  = B X H.  Art.  627. 

.*.  O-ABC,  or  7=|5X  E.  Ax.  5. 

0-  e.  d. 

Ex.  Find  the  volume  of  a triangular  pyramid  whose  altitude  is  12 
ft.,  and  whose  base  is  an  equilateral  triangle  with  a side  of  15  ft. 


PYRAMIDS 


385 


Proposition  XIX.  Theorem 

651.  The  volume  of  any  pyramid  is  equal  to  one-third 
the  product  of  its  base  by  its  altitude. 

O 


Given  the  pyramid  O-ABCDF,  having  its  volume  denoted 
by  V,  the  area  of  its  base  by  B,  and  its  altitude  by  H. 

To  prove  V=  i B X H. 

Proof.  Through  any  lateral  edge,  as  OD,  and  the  diago- 
nals of  the  base  drawn  from  its  foot,  as  AT)  and  BD,  pass 
planes  dividing  the  pyramid  into  triangular  pyramids. 

Then  V,  the  volume  of  the  pyramid  O-ABCDF , will 
equal  the  sum  of  the  volumes  of  the  triangular  pyramids. 

But  the  volume  of  each  A pyramid  = i its  base  X H. 

Art.  650. 

Hence  the  sum  of  the  volumes  of  A pyramids  — § sum  of 
their  bases  XI.  Ax.  2. 

= | j B X H.  Ax.  8. 

7=1  B X IT.  Ax.  l. 

Q.  E.  D. 

652.  Cor.  1.  The  volumes  of  two  pyramids  are  to  each 
other  as  the  products  of  their  bases  and  altitudes ; pyramids 
having  equivalent  bases  and  equal  altitudes  are  equivalent . 

653.  Cor.  2.  Pyramids  having  equivalent  bases  are  to 
each  other  as  their  altitudes;  pyramids  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 

654.  Scholium.  The  volume  of  any  polyhedron  maybe 
found  by  dividing  the  polyhedron  into  pyramids , finding  the 
volume  of  each  pyramid  separately , and  taking  their  sum. 


386 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XX.  Theorem- 

655.  The  frustum  of  a triangular  pyramid  is  equivalent 
to  the  sum  of  three  pyramids  whose  common  altitude  is  the 
altitude  of  the  frustum,  and  whose  bases  are  the  lower  base, 
the  upper  base,  and  a mean  proportional  between  the  two 
bases  of  the  frustum. 


having  the  area  of  its  lower  base  denoted  by  B,  the  area 
of  its  upper  base  by  b,  and  its  altitude  by  E. 

To  prove  ABC-DEE  three  pyramids  whose  bases  are 
B,  b and  V Bb,  and  whose  common  altitude  is  E. 

Proof.  Through  E and  AC,  E and  DC,  pass  planes  divid- 
ing the  frustum  into  three  triangular  pyramids.  Then 

1.  E-ABC  has  the  base  B and  the  altitude  E. 

2.  E-DEC,  that  is,  C-DEF,  has  the  base  b and  the 
altitude  E. 

3.  It  remains  to  show  that  E-ADG  is  equivalent  to  a 
pyramid  having  an  altitude  E,  and  a base  that  is  a mean 
proportional  between  B and  b. 

Denoting  the  three  pyramids  by  I,  II,  III, 

I A ABE  AB  = A6 _ A ADC  II 

II  A ADE  DE  DF  A DEC  III' 

(Arts.  653,  391,  644,  321.  Let  the  pupil  supply  the  reason  for  each 
step  in  detail) . 

= <Ax-  1)  Or  II  = V/IXIII.  Art.  303. 

/.  E-ADC=V/ ($EX  B)  HEX  b)=h  eV B X b. 

Hence,  ABC-DEFosum  of  three  pyramids,  as  described. 

Q.  E.  D. 


PYRAMIDS 


387 


656.  Formula  for  volume  of  frustum  of  a triangular 
pyramid.  E (B  + b + YJYbT. 


Proposition  XXI.  Theorem 

657.  The  volume  of  the  frustum  of  any  pyramid  is 
equivalent  to  the  sum  of  the  volumes  of  three  pyramids,  whose 
common  altitude  is  the  altitude  of  the  frustum,  and  whose 
bases  are  the  lower  base,  the  upper  base,  and  a mean  propor- 
tional between  the  two  bases  cf  the  frustum. 

• K T 


Given  the  frustum  of  a pyramid  Ad,  having  its  volume 
denoted  by  V,  the  area  of  its  lower  base  by  B,  of  its  upper 
base  by  b,  and  its  altitude  by  E. 

To  prove  E (B  -f  b + V Bb) . 

Proof.  Produce  the  lateral  faces  of  Ad  to  meet  in  K. 

Also  construct  a triangular  pyramid  with  base  PQR 
equivalent  to  ABCBF,  and  in  the  same  plane  with  it,  and 
with  an  altitude  equal  to  the  altitude  of  K-ABCJDF.  Pro- 
duce the  plane  of  ad  to  cut  the  second  pyramid  in  pqr. 

Then  pqroabcdf.  Art.  647. 

pyramid  E-ABCBF  ==  pyramid  T-PQB.  Art.  652. 

Also  pyramid  K-abcdf  jc=  pyramid  T-pqr.  (Why  ?) 

Subtracting,  frustum  Ad  =c=  frustum  Pr.  Ax.  3. 

But  volume  Pr  — ^E  {B  + b + Y Bb) . 

volume  Ad— $ E {B  -f  b -f-  V Bb) . (Why  ?) 

Q.  E.  D. 


388 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XXII.  Theorem 

658.  A truncated  triangular  prism  is  equivalent  to  the 
sum  of  three  pyramids,  of  which  the  base  of  the  prism  is  the 
common  base,  and  ivhose  vertices  are  the  three  vertices  of  the 
inclined  section. 


Given  the  truncated  triangular  prism  ABC-  PQR. 

To  prove  ABC-PQR  o the  sum  of  the  three  pyramids 
P-ABC,  Q-ABC  and  R-ABC. 

Proof.  Pass  planes  through  Q and  AC,  Q and  PC,  divid- 
ing the  given  figure  into  the  three  pyramids  Q-ABC, 
Q-APC  and  Q-PRC. 

1.  Q-ABC  has  the  required  base  and  the  required 
vertex  Q. 

2.  Q-APC  - B-APC,  Art.  652. 

( for  they  have  the  same  base,  APC,  and  the  same  altitude,  their  ver- 
tices being  in  a line  ||  base  APC). 

But  B-  APC  may  be  regarded  as  having  P for  its  ver- 
tex, and  ABC  for  its  base,  as  desired.  Art.  634. 

3.  Q-PRC  =0=  B-ARC  (see  Fig.  2).  Art,  652. 

( for  the  base  ABC  o base  PEC  (Art.  390);  and  the  altitudes  of  the  two 
pyramids  are  equal,  the  vertices  Q and  B being  in  line  ||  plane 
PACE,  in  which  the  bases  lie). 

But  B-ARC  may  be  regarded  as  having  R for  its  ver- 
tex, and  ABC  for  its  base,  as  desired.  Art.  634. 


PKISMATOIDS 


389 

ABC-PQB  =0=  sum  of  three  pyramids  whose  common 
base  is  ABC,  and  whose  vertices  are  P,  Q,  B. 

Q.  E.  D. 


659.  Cor.  1.  The  volume  of  a truncated  right  triangu- 
lar prism  (Fig.  3)  is  equal  to  the  product  of  its  base  by 
one-third  the  sum  of  its  lateral  edges. 


660.  Cor.  2.  The  volume  of  any  truncated  triangular 
prism  (Fig.  4)  is  equal  to  the  product  of  the  area  of  its 
right  section  by  one -third  the  sum  of  its  lateral  edges. 


PRISMATOIDS 

661.  A prismatoid  is  a polyhedron 
bounded  by  two  polygons  in  parallel  planes, 
called  bases,  and  by  lateral  faces  which  are 
either  triangles,  trapezoids  or  parallelograms. 


662.  A prismoid  is  a prismatoid  in 
which  the  bases  have  the  same  number  of 
sides  and  have  their  corresponding  sides 
parallel . 


Prismoid 


Ex.  The  volume  of  a truncated  right  parallelopiped  equals  the 
area  of  the  lower  base  multiplied  by  one-fourth  the  sum  of  the  lateral 
edges  (or  by  a perpendicular  from  the  center  of  the  upper  base  to  the 
lewer  base). 


390 


BOOK  VII.  SOLID  GEOMETKY 


Proposition  XXIII.  Theorem 

663.  The  volume  of  a prismatoid  is  equal  to  one-sixth 
the  'product  of  its  altitude  by  the  sum  of  its  bases  and  of 
four  times  the  area  of  its  midsection. 

F k 


D 


Eigt  1 

Given  the  prismatoid  ABCD-FGK,  with  bases  B and  h, 
midsection  M,  volume  V,  and  altitude  E. 

To  prove  V=i  E {B  + b + 4 M). 

Proof.  Take  any  point  0 in  the  midsection,  and  through 
it  and  each  edge  of  the  prismatoid  let  planes  be  passed. 
These  planes  will  divide  the  figure  into  pai’ts  as  follows: 

1.  A pyramid  with  vertex  0,  base  ABCD  and  altitude 

£ E,  and  whose  volume  = } E X B.  Art.  651. 

2.  A pyramid  with  vertex  0,  base  FGK,  and  altitude  J 

E , and  whose  volume  =i  EX  b.  (Why?) 

3.  Tetrahedrons  like  O-ABG  whose  volume  may  be 
* determined  as  follows  (see  Fig.  2): 

AB  = 2 PQ.  (Why?) 

/.  A AGB  = 4:  A PGQ.  Art.39S. 

.*.  0-AGB  = 4:  O-PGQ.  Art.  653. 

But  O-PGQ  (or  G-PQO)  = h PQO  Xh  E = \ EX  PQO. 

0-AGB=i  EX  4 A PQO.  (Why?) 
.*.  the  sum  of  all  tetrahedrons  like  0~AGB  = \ FXil. 

.*.  V=\  EXB  + iEXb  + iEXiM. 

Or  V=h  E (B+  & + 4 10.  q.  e.  D. 


G? 


B 

Fig.  2 


REGULAR  POLYHEDRONS 


391 


REGULAR  POLYHEDRONS 

664.  Def.  A regular  polyhedron  is  a polyhedron  all  of 
whose  faces  are  equal  regular  polygons,  and  all  of  whose 
polyhedral  angles  are  equal.  Thus,  the  cube  is  a regular 
polyhedron. 

^ Proposition  XXIV.  Theorem 

665.  But  five  regular  polyhedrons  are  possible. 


Given  regular  polygons  of  3,  4,  5,  etc.,  sides. 

To  prove  that  regular  polygons  of  the  same  number  of 
sides  can  be  joined  to  form  polyhedral  i of  a regular 
polyhedron  in  but  five  different  ways,  and  that,  conse- 
quently, but  five  regular  polyhedrons  are  possible. 

Proof.  The  sum  of  the  face  A of  any  polyhedral  angle 
< 360°.  Art.  583. 

1.  Each  Z of  an  equilateral  triangle  is  60°.  Art.  134. 
3 X 60°,  4 X 60°  and  5 X 60°  are  each  less  than  360°; 

but  any  larger  multiple  of  60°  = or  > 360°. 

.\  but  three  regular  polyhedrons  can  be  formed  with 
equilateral  A as  faces. 

2.  Each  Z of  a square  contains  90°.  Art.  151. 

3 X 90°  is  less  than  360°,  but  any  larger  multiple  of 

90°  = or  > 360°. 

but  one  regular  polyhedron  can  be  formed  with 
squares  as  faces. 

3.  Each  Z of  a regular  pentagon  is  108°.  Art.  174. 

3 X 108°  is  less  than  360°,  but  any  larger  multiple  of 

108°  > 360°. 


392 


BOOK  VII.  SOLID  GEOMETEY 


but  one  regular  polyhedron  can  be  formed  with  regu- 
lar pentagons  as  faces. 

4.  Each  Z of  a regular  hexagon  is  120°,  and  3 X 120° 
-360°. 

no  regular  polyhedron  can  be  formed  with  hexagons, 
or  with  polygons  with  a greater  number  of  sides  as  faces, 
but  five  regular  polyhedrons  are  possible. 

Q.  E.  D. 

666.  The  construction  of  the  regular  polyhedrons,  by  the 

use  of  cardboard,  may  be  effected  as  follows: 

Draw  on  a piece  of  cardboard  the  diagrams  given  below. 
Cut  the  cardboard  half  through  at  the  dotted  lines  and  en- 
tirely through  at  the  full  lines.  Bring  the  free  edges 
together  and  keep  them  in  their  respective  positions  by 
some  means,  such  as  pasting  strips  of  paper  over  them. 


POLYHEDRONS 


393 


POLYHEDRONS  IN  GENERAL 

^Proposition  XXV.  Theorem 

667.  In  any  polyhedron , the  number  of  edges  increased 
by  two  equals  the  number  of  vertices  increased  by  the  number 


Given  the  polyhedron  AT,  with  the  number  of  its  ver- 
tices, edges  and  faces  denoted  by  V,  E and  F,  respectively. 

To  prove  E + 2 = V + F. 

Proof.  Taking  the  single  face  ABCD,  the  number  of 
edges  equals  the  number  of  vertices,  or  E=  V. 

If  another  face,  CRTD,  be  annexed  (Fig.  2),  three  new 
edges,  CB,  BT,  TE,  are  added  and  two  new  vertices,  B and  T. 

the  number  of  edges  gains  one  on  the  number  of  ver- 
tices, or  E—  F+  1. 

If  still  another  face,  BQBC,  be  annexed,  two  new  edges, 
BQ  and  QB,  are  added,  and  one  new  vertex,  Q.  E—  V+2. 

With  each  new  face  that  is  aunexed,  the  number  of 
edges  gains  one  on  the  number  of  vertices,  till  but  one 
face  is  lacking. 

The  last  face  increases  neither  the  number  of  edges  nor 
of  vertices. 

Hence  number  of  edges  gains  one  on  number  of  vertices, 
for  every  face  except  two,  the  first  and  the  last,  or  gains 
F — 2 in  all. 

.\  for  the  entire  figure,  E=  V + F — 2. 

That  is  -Z£  + 2—  V + F.  Ax.  2. 

Q.  E.  D. 


394 


BOOK  VII.  SOLID  GEOMETEY 


y Proposition  XXVI.  Theorem 

668.  The  sum  of  the  face  angles  of  any  polyhedron 
equals  four  right  angles  taken  as  many  times , less  two,  as 
the  polyhedron  has  vertices. 


Given  any  polyhedron,  with  the  sum  of  its  face  angles 
denoted  by  S,  and  the  number  of  its  vertices,  edges  and 
faces  deuoted  by  F,  E,  F,  respectively. 

To  prove  S=  ( V — 2)  4 rt.  A . 

Proof.  Each  edge  of  the  polyhedron  is  the  intersection 
of  two  faces,  .'.  the  number  of  sides  of  the  faces  — 2 E. 

the  sum  of  the  interior  and  exterior  A of  the  faces  = 
2 E X 2 rt.  A , or  E X 4 rt.  A . Art.  73. 

But  the  sum  of  the  exterior  A of  each  face  = 4 rt.  A . 

Art.  175. 

.'.  the  sum  of  exterior  A of  the  E faces  — F X 4 rt.  A . 

Ax.  4. 

Subtracting  the  sum  of  the  exterior  A from  the  sum  of 
all  the  A,  the  sum  of  the  interior  A of  the  F faces  = 
(TX  4 rt.  A)  — {EX  4 rt.  A). 

Or  S=  (E — F)  4 rt.  A . 

But  E 2 = F -f-  F.  Art.  667. 

Hence  E — F—V — 2.  Ax.  3. 

Substituting  for  E — F,  S—  ( F— 2)  4 rt.  A . Ax.  8. 

Q.  E.  D, 


Ex.  Verify  the  last  two  theorems  in  the  ease  of  the  cube. 


COMPARISON  OF  POLYHEDRONS 


395 


COMPARISON  OF  POLYHEDRONS.  SIMILAR  POLYHEDRONS 

Proposition  XXVII.  Theorem 

l 669.  If  tivo  tetrahedrons  have  a trihedral  angle  of  one 
equal  to  a trihedral  angle  of  the  other , they  are  to  each  other 
as  the  products  of  the  edges  including  the  equal  trihedral 
angles. 


their  volumes  denoted  by  Y and  Y' , respectively,  and  having 
the  trihedral  A 0 and  O'  equal. 

Y OA  X OBXOC 

lo  prove  v,  0,A,  x 0,B,  x 0/(7- 

Proof.  Apply  the  tetrahedron  O'-A'B'C  to  O-ABC  so 
that  the  trihedral  Z O'  shall  coincide  with  its  equal,  the 
trihedral  Z 0. 

Draw  CP  and  C'P'  J_  plane  OAB,  and  draw  OP  the 
projection  of  OC  in  the  plane  OAB. 

Taking  OAB  and  OA'B'  as  the  bases,  and  CP  and  C'P ' 
as  the  altitudes  of  the  pyramids  C-OAB  and  C'-OA'B 
. respectively. 

Y A OAB  X CP 

Y 


But 


A OA'B1  X C’P' 

A OAB  OA  X OB 


A OAB  CP 
A OA'B'  C’P'' 


Art.  652. 


Y_ 

V1 


A OA'B'  OA'  X OB' 

CP  _ OG 
CP’  ocr 
OAXOBX  OC 


In  the  similar  rt.  A 0(7P  and  OC'P', 
OAX  OB  X OC 


Art.  397. 


(Why  ?) 


OA'  X OB'  X OC' 


O'  A'  X O'B'  X O'C 

Q.  E.  D 


Ax.  8. 


396 


BOOK  VII.  SOLID  GEOMETEY 


670.  Def.  Similar  polyhedrons  are  polyhedrons  having 
the  same  number  of  faces,  similar,  each  to  each,  and  simi- 
larly placed,  and  having  their  corresponding  polyhedral 
angles  equal. 

Proposition  XXVIII.  Theorem 

671.  Any  two  similar  polyhedrons  may  be  decomposed 
into  the  same  number  of  tetrahedrons,  similar,  each  to  each, 
and  similarly  placed. 


L 


Given  P and  P7,  two  similar  polyhedrons. 

To  prove  that  P and  P’  may  be  decomposed  into  the 
same  number  of  tetrahedrons,  similar,  each  to  each. 

Proof.  Take  R and  R'  any  two  homologous  vertices  of 
P and  P'.  Draw  homologous  diagonals  in  all  the  faces  of 
P and  P'  except  those  faces  which  meet  at  R and  R , sepa- 
rating the  faces  into  corresponding  similar  triangles. 

Through  R and  each  face  diagonal  thus  formed  in  P, 
and  through  R and  each  face  diagonal  in  P7,  pass  planes. 

Each  corresponding  pair  of  tetrahedrons  thus  formed 
may  be  proved  similar. 

Thus,  in  R-ABG  and  R-A'B'C,  the  A RBA  and 
R'B'A'  are  similar.  Art.  329. 

In  like  manner  A RBG  and  RB'C'  are  similar;  and  A 
ABG  and  A'B'G'  are  similar. 


SIMILAR  POLYGONS 


397 


Also 


HA 

H'A' 


HG  _/  BO  \ 
H'C'  \B'C'J 


AG 

A'Gr 


Art.  321. 


.'.  A AHG  and  A' H'C'  are  similar.  Art.  326. 
Hence  the  corresponding  faces  of  H-ABG  and  H'-A'B'G' 
are  similar. 

Also  their  homologous  trihedral  A are  equal.  Art.  584. 

tetrahedron  H-ABG  is  similar  to  H'-A'B'C'.  Art.  670. 
After  removing  H-ABG  from  P,  and  H'-A'B'C'  from 
P',  the  remaining  polyhedrons  are  similar,  for  their  faces 

are  similar,  and  the  remaining  polyhedral  A are  equal. 

Ax.  3. 

By  continuing  this  process,  P and  P'  may  be  decom- 
posed into  the  same  number  of  tetrahedrons,  similar,  each 
to  each,  and  similarly  placed. 

Q.  E.  D. 


672.  Cor.  1.  The  homologous  edges  of  similar  polyhe- 
drons are  proportional ; 

Any  ttvo  homologous  lines  in  two  similar  polyhedrons 
have  the  same  ratio  as  any  other  two  homologous  lines „ 


673.  Cor.  2.  Any  two  homologous  faces  of  two  similar 
polyhedrons  are  to  each  other  as  the  squares  of  any  two 
homologous  edges  or  lines; 

The  total  areas  of  any  two  similar  polyhedrons  are  to 
each  other  as  the  squares  of  any  two  homologous  edges. 


Ex.  1.  In  the  figure,  p.  395,  if  the  edges  meeting  at  G are  8,  9,  12  in., 
and  those  meeting  at  O'  are  4,  6,  8 in.,  find  the  ratio  of  the  volumes 
of  the  tetrahedrons. 

Ex.  2.  If  the  linear  dimensions  of  one  room  are  twice  as  great  as 
the  corresponding  dimensions  of  another  room,  how  will  their  surfaces 
(and  .'.  cost  of  papering)  compare  ? How  will  their  volumes  compare  ? 

Ex.  3.  How  many  2 in.  cubes  can  be  cut  from  a 10  in.  cube  ? 

Ex.  4.  If  the  bases  of  a prismoid  are  rectangles  whose  dimensions 
are  a,  b and  b,  a,  and  altitude  is  H,  find  the  formula  for  the  volume. 


oy» 


BOOK  VII.  SOLID  GEOMETRY 


Proposition  XXIX.  Theorem 

674.  The  volumes  of  two  similar  tetrahedrons  are  to 
each,  other  as  the  cubes  of  any  pair  of  homologous  edges. 


Given  the  similar  tetrahedrons  O-ABC  and  O'-A'B'C. 
Y 1 XA3 


To  prove 


O'A'3 


Proof. 


But 


Y OA  X OB  X 00 
V'  O'A'  X O'B’  X O' O' 

_ OA  y OB  y 00 
O'A'  O'B'  O' C'' 

OA  _ OB  _ PC 
O'A'  O'B'  O' a' 

. V 0A  X 0A  ><  0A  - ~^~3 

■■  Yr  O'A'  O'A'  O'A'  OhP3 


(Why  f) 


Art.  673. 


Ax.  8. 


Ex.  1.  In  the  above  figures,  if  AB=  2 A'B',  find  the  ratio  of  V 
to  V.  Find  the  same,  if  AB= li  A'B'.  < 

Ex.  2 The  measurement  of  the  volume  of  a regular  triangular 
prism  reduces  to  the  measurement  of  the  lengths  of  how  many  straight 
lines  ? of  a frustum  of  a regular  square  pyramid  ? 

Ex.  3.  Show  how  to  construct  out  of  pasteboard  a regular  prism, 
a parallelopiped,  and  a truncated  square  prism. 


SIMILAR  POLYGONS 


399 


Proposition  XXX.  Theorem 


675.  The  volumes  of  any  two  similar  polyhedrons  are 
to  each  other  as  the  cubes  of  any  two  homologous  edges , or 
of  any  other  two  homologous  lines. 


L 


Given  the  polyhedrons  AK  and  A'K!  having  their  vol- 
umes denoted  by  V and  V , and  RB  and  R'B'  any  pair  of 
homologous  edges. 


To  prove 


V _ HB3 

V R'B'3 


Proof.  Let  the  polyhedrons  be  decomposed  into  tetra- 
hedrons, similar,  each  to  each,  and  similarly  placed.  Art.  671. 

Denote  the  volumes  of  the  tetrahedrons  in  P by  Vi, 

Vi  . . . and  of  those  in  P ' by  v\,  v'z,  v'z  . . . 


Then 

Also 


HB 


vi_ 

vf  'eJb'3 

Vi  V 2 t'3  

V\  Vr2  V'z 


Art.  674,  Ax.  1. 


HB6 


( for  each  of  these  rati  os  = -••) 

R'B'3 


. ^1  ~h  V2  ~l~  t'3  ~t~  — _ t'i  ,,  , . y _Vi 

••  «/l  + ®,2  + «,3  + — “«'i;  ’ V'  *'■ 


Y_ 

v' 


v 1 

_ HB3 

Wb1' 


v’i 


Art.  312. 


Ax.  1 


Q.  E.  D- 


400 


BOOK  VII.  SOLID  GEOMETRY 


EXERCISES.  CROUP  67 

THEOREMS  CONCERNING  POLYHEDRONS 

Ex.  1.  The  lateral  faces  of  a right  prism  are  rectangles 

Ex.  2.  A diagonal  plane  of  a prism  is  parallel  to  every  lateral  edge 
of  the  prism  not  contained  in  the  plane. 

Ex.  3.  The  diagonals  of  a parallelopiped  bisect  each  other. 

Ex.  4.  The  square  of  a diagonal  of  a rectangular  parallelopiped 
equals  the  sum  of  the  squares  of  the  three  edges  meeting  at  a vertex. 

Ex.  5.  Each  lateral  face  of  a prism  is  parallel  to  every  lateral  edge 
not  contained  in  the  face. 

Ex.  6.  Every  section  of  a prism  made  by  a plane  parallel  to  a 
lateral  edge  is  a parallelogram. 

Ex.  7.  If  any  two  diagonal  planes  of  a prism  which  are  not  par' 
allel  to  each  other  are  perpendicular  to  the  base  of  the  prism,  the 
prism  is  a right  prism. 

Ex.  8.  What  part  of  the  volume  of  a cube  is  the  pyramid  whose 
base  is  a face  of  the  cube  and  whose  vertex  is  the  center  of  the  cube  ? 

Ex,  9.  Any  section  of  a regular  square  pyramid  made  by  a plane 
through  the  axis  is  an  isosceles  triangle. 

Ex.  10.  In  any  regular  tetrahedron,  an  altitude  equals  three  times 
the  perpendicular  from  its  foot  to  any  face. 

Ex.  11.  In  any  regula'r  tetrahedron,  an  altitude  equals  the  sum  of 
the  perpendiculars  to  the  faces  from  any  point  within  the  tetrahedron 

Ex.  12.  Find  the  simplest  formula  for  the  lateral  area  of  a trun- 
cated regular  prism  of  n sides. 

Ex.  13.  The  sum  of  the  squares  of  the  four  diagonals  of  a paral- 
lelopiped is  equal  to  the  sum  of  the  squares  of  the  twelve  edges. 

[Sue.  Use  Art.  352.] 

Ex-  14.  A parallelopiped  is  symmetrical  with  respect  to  what  point? 

Ex.  15.  A rectanglar  parallelopiped  is  symmetrical  with  respect  to 
how  many  planes?  (Let  the  pupil  make  a definition  of  a figure  sym- 
metrical with  respect  to  a plane.  See  Arts.  486,  487.) 


EXERCISES  ON  POLYHEDRONS 


401 


Ex.  16.  The  volume  of  a pyramid  whose  lateral  edges  are  the 
three  edges  of  the  parallelopiped  meeting  at  a point  is  what  part  of 
the  volume  of  the  parallelopiped  1 

Ex.  It  If  a plane  be  passed  through  a vertex  of  a cube  and  the 
diagonal  of  a face  not  adjacent  to  the  vertex,  what  part  of  the  volume 
of  the  cube  is  contained  by  the  pyramid  so  formed  ? 

Ex.  18.  If  the  angles  at  the  vertex  of  a triangular  pyramid  are 
right  angles  and  each  lateral  edge  equals  a,  show  that  the  volume  of 

the  pyramid  is  -jr' 

Ex.  19.  How  large  is  a dihedral  angle  at  the  base  of  a regular 
■pyramid,  if  the  apothem  of  the  base  equals  the  altitude  of  the  pyramid  ? 

Ex.  20.  The  area  of  the  base  of  a pyramid  is  less  than  the  area  of 
the  lateral  surface. 


Ex.  21.  The  section  of  a triangular  pyramid  by  a plane  parallel 
to  two  opposite  edges  is  a parallelogram. 

If  the  pyramid  is  regular,  what  kind  of  a parallelogram  does  the 
section  become  ? 

Ex.  22.  The  altitude  of  a regular  tetrahedron  divides  an  altitude 
of  the  base  into  segments  which  are  as  2 : 1. 


Ex.  23.  If  the  edge  of  a regular  tetrahedron  is  a,  show  that  the 
slant  height  is  ; and  hence  that  the  altitude  is  and  the  vol- 

O 


. rtV  2 

ume  is  — — - 

Ex.  24.  If  the  midpoints  of  all  the  edges  of 
a tetrahedron  except  two  opposite  edges  be 
joined,  a parallelogram  is  formed. 

Ex.  25.  Straight  lines  joining  the  midpoints 
of  the  opposite  edges  of  a tetrahedron  meet  in 
a point  and  bisect  each  other. 


Ex.  26.  The  midpoints  of  the  edges  of  a regular  tetrahedron  are 
the  vertices  of  a regular  octahedron. 


Z 


402 


BOOK  VII.  SOLID  GEOMETRY 


EXERCISES.  CROUP  68 

PROBLEMS  CONCERNING  POLYHEDRONS 

iJ-x.  1 . Bisect  the  volume  of  a given  prism  by  a plane  parallel  to 
the  base. 

Ex.  2.  Bisect  the  lateral  surface  of  a given  pyramid  by  a plane  par- 
allel  to  the  base. 

Ex.  3.  Through  a given  point  pass  a plane  which  shall  bisect  the 
volume  of  a given  parallelopiped. 

Ex.  4.  Given  an  edge,  construct  a regular  tetrahedron. 

Ex.  5.  Given  an  edge,  construct  a regular  octahedron. 

Ex.  6.  Pass  a plane  through  the  axis  of  a regular  tetrahedron  so 
that  the  section  shall  be  an  isosceles  triangle. 

Ex.  7.  Pass  a plane  through  a cube  so  that  the  section  shall  be  a 
regular  hexagon. 

Ex.  8.  Through  three  given  lines  no  two  of  which  are  parallel 
pass  planes  which  shall  form  a parallelopiped. 

Ex.  9.  Prom  cardboard  construct  a regular  square  pyramid  each 
of  whose  faces  is  an  equilateral  triangle. 

EXERCISES.  CROUP  69 


REVIEW  EXERCISES 
Make  a list  of  the  properties  of 


Ex.  1. 

Straight  lines  in  space. 

Ex.  9.  Right  prisms. 

Ex.  2. 

One  line  and  one  plane. 

Ex.  10.  Parallelopipeds  in  gen- 

Ex. 3. 

Two  or  more  lines  and 

eral. 

one 

plane . 

Ex.  11.  Rectangular  parallelo- 

Ex. 4. 

Two  planes  and  one  line. 

pipeds. 

Ex.  5. 

Two  planes  and  two  lines. 

Ex.  12.  Pyramids  in  general. 

Ex.  6. 

Polyhedrons  in  general. 

Ex.  13.  Regular  pyramids. 

Ex.  7. 

Similar  polyhedrons. 

Ex.  14  Frusta  of  pyramids. 

Ex.  8. 

Prisms  in  general. 

Ex.  15.  Truncated  prisms. 

Book  VIII 


CYLINDERS  AND  CONES 

CYLINDERS 

676.  A cylindrical  surface  is 

a curved  surface  generated  by  a 
straight  line  which  moves  so  as 
constantly  to  touch  a given  fixed 
curve  and  constantly  be  parallel 
to  a given  fixed  straight  line. 

Thus,  every  shadow  cast  by  a point 
of  light  at  a great  distance,  as  by  a 
star  or  the  sun,  approximates  the 
cylindrical  form,  that  is,  is  bounded 
by  a cylindrical  surface  of  light.  Hence,  in  all  radiations  (as  of  light, 
heat,  magnetism,  etc.)  from  a point  at  a great  distance,  we  are 
concerned  with  cylindrical  surfaces  and  solids. 

677.  The  generatrix  of  a cylindrical  surface  is  the  mov- 
ing straight  line;  the  directrix  is  the  given  curve,  as  CDE; 
an  element  of  the  cylindrical  surface  is  the  moving  straight 
line  in  any  one  of  its  positions,  as  DF. 

678.  A cylinder  is  a solid  bounded  by 
a cylindrical  surface  and  by  two  parallel 
planes. 

The  bases  of  a cylinder  are  its  parallel 
plane  faces;  the  lateral  surface  is  the 
cylindrical  surface  included  between  the 
parallel  planes  forming  its  bases;  the  alti- 
tude of  a cylinder  is  the  distance  between  the  bases. 

The  elements  of  a cylinder  are  the  elements  of  the 
cylindrical  surface  bounding  it. 

(403) 


404: 


BOOK  VIII.  SOLID  GEOMETRY 


679.  Property  of  a cylinder  inferred  immediately.  AU 

the  elements  of  a cylinder  are  equal,  for  they  are  parallel  lines 
included  between  parallel  planes  (Arts.  532,  676). 

Tlie  cylinders  most  important  in  practical  life  are  those  determined 
by  their  stability,  the  ease  with  which  they  can  be  made  from  com- 
mon materials,  etc. 

680.  A right  cylinder  is  a cylinder 
whose  elements  are  perpendicular  to  the 
bases. 

681.  An  oblique  cylinder  is  one  whose 
elements  are  oblique  to  the  bases. 

682.  A circular  cylinder  is  a cylinder 
whose  bases  are  circles. 

683.  A cylinder  of  revolution  is  a cylin- 
der generated  by  the  revolution  of  a rect- 
angle about  one  of  its  sides  as  an  axis. 

Hence,  a cylinder  of  revolution  is  a right 
circular  cylinder. 

Some  of  the  properties  of  this  solid  are  derived 
most  readily  by  considering  it  as  generated  by  a re- 
volving rectangle  ; and  others,  by  regarding  it  as  a 
particular  kind  of  cylinder  derived  from  the  general 
definition. 

684.  Similar  cylinders  of  revolution  are  cylinders  gen- 
erated by  similar  rectangles  revolving  about  homologous 
sides. 

685.  A tangent  plane  to  a cylinder  is  a plane  which 
contains  one  element  of  the  cylinder,  and  which  does  not  cut 
the  cylinder  on  being  produced. 

Ex.  1.  A plane  passing  through  a tangent  to  the  base  of  a circu- 
lar cylinder  and  the  element  drawn  through  the  point  of  contact  is 
tangent  to  the  cylinder.  (For  if  it  is  not,  etc.) 

Ex.  2.  If  a plane  is  tangent  to  a circular  cylinder,  its  intersection 
■with  the  nlane  of  the  base  is  tangent  to  the  base. 


Oblique  circular 
cylinder 


Cylinder  of 
revolution 


CYLINDERS 


405 


686.  A prism  inscribed  in  a cylinder  is  a prism  whose 
lateral  edges  are  elements  of  the  cylinder,  and  whose  bases 
are  polygons  inscribed  in  the  bases  of  the  cylinder. 


687.  A prism  circumscribed  about  a cylinder  is  a prism 
whose  lateral  faces  are  tangent  to  the  cylinder,  and  whose 
bases  are  polygons  circumscribed  about  the  bases  of  the 
cylinder. 

688.  A section  of  a cylinder  is  the  figure  formed  by  the 
intersection  of  the  cylinder  by  a plane. 

A right  section  of  a cylinder  is  a section  formed  by  a 
plane  perpendicular  to  the  elements  of  the  cylinder. 

689.  Properties  of  circular  cylinders.  By  Art.  441  the 
area  of  a circle  is  the  limit  of  the  area  of  an  inscribed  or 
circumscribed  polygon,  and  the  circumference  is  the  limit 
of  the  perimeters  of  these  polygons;  hence 

1.  The  volume  of  a circular  cylinder  is  the  limit  of  the 
volume  of  an  inscribed  or  circumscribed  prism. 

2.  The  lateral  area  of  a circular  cylinder  is  the  limit  of 
the  lateral  area  of  an  inscribed  or  circumscribed  prism. 

Also,  3.  By  methods  too  advanced  for  this  hook , it  may  he  proved  that 
the  perimeter  of  a right  section  is  the  limit  of  the  perimeter  of  a right  sec- 
tion of  an  inscribed  or  circumscribed  vrism . 


Inscribed  prism 


Circumscribed  prism 


406 


BOOK  Yin.  SOLID  GEOMETRY 


Proposition  I. 


Theorem 


690.  Every  section  of  a cylinder  made  by  a plane  pass - 
ing  through  an  element  is  a parallelogram. 


Given  the  cylinder  AQ  cut  by  a plane  passing  through 
the  element  AB  and  forming  the  section  ABQP. 

To  prove  ABQP  a CD  . 

Proof.  AP  ||  BQ.  Art.  531. 

It  remains  to  prove  that  PQ  is  a straight  line  1 1 AB. 

Through  P draw  a line  in  the  cutting  plane  ||  AB. 

This  line  will  also  lie  in  the  cylindrical  surface.  Art.  676. 

this  line  must  coincide  with  PQ, 

( for  the  line  drawn  lies  in  both  the  cutting  plane  and  the  cylindrical 
surface,  hence,  it  must  be  their  intersection) . 

PQ  is  a straight  line  ||  AB. 

:.  ABQP  is  a CD  . (Why  ?) 

Q.  E.  D. 

691.  Cor.  Every  section  of  a right  cylinder  made  ly  a 
plane  passing  through  an  element  is  a rectangle. 

Ex.  1.  A door  swinging  on  its  hinges  generates  what  kind  of  a 
solid  ? 

Ex.  2.  Every  section  of  a parallelopiped  made  by  a plane  inter- 
secting all  its  lateral  edges  is  a parallelogram. 


CYLINDERS 


407 


Given  the  cylinder  AQ  with  the  bases  APB  and  GQD. 
To  prove  base  APB  = base  CQD. 

Proof.  Let  AC  and  BT>  be  any  two  fixed  elements  in 
the  surface  of  the  cylinder  AQ. 

Take  P,  any  point  except  A and  B in  the  perimeter  of 


the  base,  and  through  it  draw  the  element  PQ. 

Draw  AB,  AP,  PB,  CD,  CQ,  QD. 

Then  AC  and  BD  are  — and  ||.  (Why?) 

AD  is  a LU  . (Why?) 

Similarly  AQ  and  BQ  are  ZjE7  . 

AB=CD,  AP=CQ,  and  BP=DQ.  (Why?) 

A APB  = A CQD.  (Why  ?) 


Apply  the  base  APB  to  the  base  CQD  so  that  AB  coin- 
cides with  CD.  Then  P will  coincide  with  Q, 

(for  A APB  = A CQD). 

But  P is  any  point  in  the  perimeter  of  the  base  APB. 

:.  every  point  in  the  perimeter  of  the  lower  base  will 
coincide  with  a corresponding  point  of  the  perimeter  of 
die  upper  base. 

the  bases  will  coincide  and  are  equal.  Art.  47. 

E. 


408 


BOOK  Yin.  SOLID  GEOMETRY 


693.  Cor.  1.  The  sections  of  a cylinder  made  by  two 
parallel  planes  cutting  all  the  elements  are  equal. 

For  the  sections  thus  formed  are  the  bases  of  the  cylinder 
included  between  the  cutting  planes. 

694.  Cor.  2.  Any  section  of  a cylinder  parallel  to  the 
bast  ' kal  to  the  base. 


695.  The  lateral  area  of  a circular  cylinder  is  equal  to 
the  product  of  the  perimeter  of  a right  section  of  the  cylin- 
der by  an  element. 


Given  the  circular  cylinder  AJ,  having  its  lateral  area 
denoted  by  S,  an  element  by  E,  and  the  perimeter  of  a 
right  section  by  P. 


Proof.  Let  a prism  with  a regular  polygon  for  its  base 
be  inscribed  in  the  cylinder. 

Denote  the  lateral  area  of  the  inscribed  prism  by  S', 
and  the  perimeter  of  its  right  section  by  P'. 

Then  the  lateral  edge  of  the  inscribed  prism  is  an  ele- 
ment of  the  cylinder.  Constr. 


Proposition  III.  Theorem 


j 


To  prove 


S=PXE. 


CYLINDERS 


409 


S'  = P'XE. 


Art.  608. 


If  the  number  of  lateral  faces  of  the  inscribed  prism  be 


indefinitely  increased, 

S'  will  approach  S as  a limit.  Art.  689  2. 

P'  will  approach  P as  a limit.  Art.  689,  3. 

And  P'X-E/will  approach  PXE  as  a limit.  A.rt.  253,  2. 

But  S'  = P'XE  always.  (Why?) 

.*.  S—PXE.  (Why?) 


Q.  E.  1). 


696.  Cor.  1.  The  lateral  area  of  a cylinder  of  revolu- 
tion is  equal  to  the  product  of  the  circumference  of  its  base 
by  its  altitude. 


697.  Formulas  for  lateral  area  and  total  area  of  a cylin- 
der of  revolution.  Denoting  the  lateral  area  of  a cylinder  of 
revolution  by  S,  the  total  area  by  T,  the  radius  by  R,  and 
the  altitude  by  H. 

S = 2 tiRE. 

T=  2 nRH  + 2 nk2  T=  2 nR  (JT+  R). 


Ex.  1.  If,  in  a cylinder  of  revolution,  11=10  in.  and  R= 7 in., 
find  <S  and  T. 

• 

Ex.  2.  If  the  altitude  of  a cylinder  of  revolution  equals  the  radius 
of  the  base  ( H = R ),  what  do  the  formulas  for  S and  T become  in 
terms  of  Ii  ? also,  in  terms  of  H i 

Ex.  3.  What  do  they  become,  if  the  altitude  equals  the  diameter 
of  the  base  ? 

Ex.  4.  In  a cylinder  of  revolution,  what  is  the  ratio  of  the  lateral 
area  to  the  area  of  the  base  ? to  the  total  area  ? 


410 


BOOK  VIII.  SOLID  GEOMETRY 


Proposition  IV.  Theorem 

698.  The  volume  of  a circular  cylinder  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given  the  circular  cylinder  AJ,  having  its  volume 
denoted  by  V,  its  base  by  B,  and  its  altitude  by  R. 

To  prove  Y=BX  H. 

Proof.  Let  a prism  having  a regular  polygon  for  its 
base  be  inscribed  in  the  cylinder,  and  denote  the  volume  of 
the  inscribed  prism  by  V',  and  its  base  by  B'. 

The  prism  will  have  the  same  altitude,  R,  as  the 


cylinder. 

/.  V'  = B'XR.  (Why?) 

If  the  number  of  lateral  faces  of  the  inscribed  prism 
be  indefinitely*  increased, 

Y will  approach  V as  a limit.  Art.  689,  l. 
B’  will  approach  B as  a limit.  (Why?) 

And  B'  X R will  approach  B X R as  a limit  (Why  ?) 

But  V'=Bf  X R always.  (Why?) 

.*.  V=B  X R.  (Why?) 

Q.  E.  D. 


699.  Formula  for  the  volume  of  a circular  cylinder. 

By  use  of  Art.  450, 


V=nR2R, 


CYLINDERS 


411 


Proposition  V. 


Theorem 


700.  The  lateral  areas,  or  the  total  areas , of  two  simi- 
lar cylinders  of  revolution  are  to  each  other  as  the  squares 
of  their  radii,  or  as  the  squares  of  their  altitudes;  and 
their  volumes  are  to  each  other  as  the  cubes  of  their  radii , 
or  as  the  cubes  of  their  altitudes. 


Given  two  similar  cylinders  of  revolution  having  their 
lateral  areas  denoted  by  S and  S',  their  total  areas  by  T 
and  T , their  volumes  by  V and  V',  their  radii  by  B and 
B',  and  their  altitudes  by  R and  R , respectively. 

To  prove  8 : S'=T  : T'=B2  : B,2--=R2  : R2-, 
and  V:  V'  = B3  : B'3  = R3  : H'3. 


„ * R _B  _H±B 

Proof.  ]]f  R/  E,  + Rf 


Arts.  321,  309. 


8 2 7tBH  _BXR  _B  R _B2 

V 7?'  V 77/  7?/ X 77/  7?/2 ' 


R 2 


S'  2 n B'W  B'  X R'  B' 


R'  B< 
R+B 


(Why?) 


T 2 7tB(H-fB ) B 
A1S°  T'  2 7i B'(R-fB')  B'  R'-fB' 


V 7tB2R  Br  R B3  R3 
AIso  V'  7iB/2R'  b>-  R B'3  r3‘ 


R2 
B^  = IP_ 

B’2  R 2 

(Why?) 

(Why?) 

Q.  E.  D. 


Ex,  If  a cylindrical  cistern  is  12  ft.  deep,  how  much  more  cement 
is  required  to  line  it  than  to  line  a similar  cistern  6 ft.  deep  ? How 
much  more  water  will  the  former  cistern  hold  ? 


412 


BOOK  VIII.  SOLID  GEOMETRY 


CONES 


i 


B 


701.  A conical  surface  is  a sur- 
face generated  by  a straight  line 
•which  moves  so  as  constantly  to  touch 
a given  fixed  curve,  and  constantly 
pass  through  a given  fixed  point. 

Thus  every  shadow  east  by  a near  point 
of  light  is  conical  in  form,  that  is,  is  bounded 
by  a conical  surface  of  light.  Renee,  the 
study  of  conical  surfaces  and  solids  is  im- 
portant from  the  fact  that  it  concerns  all 
cases  of  forces  radiating  from  a near  point. 


702.  The  generatrix  of  a conical 
surface  is  the  moving  straight  line,  as 
AA' \ the  directrix  is  the  given  fixed 
curve,  as  ABC-,  the  vertex  is  the  fixed  point,  as  0;  an 
element  is  the  generatrix  in  any  one  of  its  positions,  as  BB' . 


703.  The  upper  and  lower  nappes  of  a conical  surface 
are  the  portions  above  and  below  the  vertex,  respectively, 
as  O-ABC  and  O-A'B'C. 

Usually  it  is  convenient  to  limit  a conical  surface  to  a single  nappe- 


704.  A cone  is  a solid  bounded  by  a 
conical  surface  and  a plane  cutting  all  the 
elements. 

705.  The  base  of  a cone  is  the  face 
formed  by  the  cutting  plane;  the  lateral 
surface  is  the  bounding  conical  surface; 
the  vertex  of  the  cone  is  the  vertex  of  the 
conical  surface;  the  elements  of  the  cone 
are  the  elements  of  the  conical  surface; 


Oblique  circular 
cone 


the  altitude  of  a cone  is  the  perpendicular  distance  from  the 


vertex  to  the  plane  of  the  base. 


CONES 


413 


706.  A circular  cone  is  a cone  whose  base  is  a circle. 
The  axis  of  a circular  cone  is  the  line  drawn  from  the 
vertex  to  the  center  of  the  base. 

707.  A right  circular  cone  is  a circular  cone  whose  axis 
is  perpendicular  to  the  plane  of  the  base. 

An  oblique  circular  cone  is  a circular  cone 
whose  axis  is  oblique  to  the  base. 

708.  A cone  of  revolution  is  a cone  gene- 
rated by  the  revolution  of  a right  triangle 
about  one  of  its  legs  as  an  axis. 

Hence  a cone  of  revolution  and  a right 
circular  cone  are  the  same  solid.  Cone  of  revolution 

709.  Properties  of  a cone  of  revolution  inferred  im- 
mediately. 

1.  The  altitude  of  a cone  of  revolution  is  the  axis  of  the 
cone. 

2.  All  the  elements  of  a cone  of  revolution  are  equal. 

710.  The  slant  height  of  a cone  of  revolution  is  any  one 

of  its  elements. 

711.  Similar  cones  of  revolution  are  cones  generated  by 
similar  right  triangles  revolving  about  homologous  sides. 

. 12.  A plane  tangent  to  a cone  is  a plane  which  con- 
tains one  element  of  the  cone,  but  which  does  not  cut  the 
conical  surface  on  being  produced. 

Ex.  1.  A plane  passing  through  a tangent  to  the  base  of  a circular 
cone  and  the  element  drawn  through  the  point  of  contact  is  tangent  to 
the  cone. 

Ex.  2.  If  a plane  is  tangent  to  a circular  cone,  its  intersection  with 
the  plane  of  the  base  is  tangent  to  the  cone. 


414 


BOOK  VIII.  SOLID  GEOMETKY 


713.  A pyramid  inscribed  in 
a cone  is  a pyramid  whose  lateral 
edges  are  elements  of  the  cone 
and  whose  base  is  a polygon  in- 
scribed in  the  base  of  the  cone. 


714.  A pyramid  circum- 
scribed about  a cone  is  a pyra- 
mid whose  lateral  faces  are  tan 
gent  to  the  cone  and  whose  base  is  a polygon  circumscribed 
about  the  base  of  the  cone. 


715.  Properties  of  circular  cones.  By  Art.  441  the  area 

of  a circle  is  the  limit  of  the  area  of  an  inscribed,  or  of  a 
circumscribed  polygon,  and  the  circumference  is  the  limit 
of  the  perimeters  of  these  polygons;  hence 

1.  The  volume  of  a circular  cone  is  the  limit  of  the  vol- 
ume of  an  inscribed  or  circumscribed  pyramid. 

2.  The  lateral  area  of  a circular  cone  is  the  limit  of  the 
lateral  area  of  an  inscribed  or  circumscribed  pyramid. 


716.  A frustum  of  a cone  is  the  por- 
tion of  the  cone  included  between  the  base 
of  the  cone  and  a plane  parallel  to  the 
base. 

The  lower  base  of  the  frustum  is  the 
base  of  the  cone,  and  the  upper  base  of 
the  frustum  is  the  section  made  by  the 
plane  parallel  to  the  base  of  the  cone. 


What  must  be  the  altitude  and  the  lateral  surface  of  a 
frustum  of  a cone;  also  the  slant  height  of  the  frustum  of 
a,  cone  of  revolution  1 


CONES 


415 


Proposition  VI.  Theorem 

717.  Every  section  of  a cone  made  by  a plane  passing 
through  its  vertex  is  a triangle. 


Given  the  cone  S-APBQ  with  a plane  passing  through 
the  vertex  S,  and  making  the  section  SPQ. 

To  prove  SPQ  a triangle. 

Proof.  PQ,  the  intersection  of  the  base  and  the  cutting 
plane,  is  a straight  line.  (Why?) 

Draw  the  straight  lines  SP  and  SQ. 

Then  SP  and  SQ  must  be  in  the  cutting  plane;  Art.  498. 

And  be  elements  of  the  conical  surface.  Art.  701. 

the  straight  lines  SP  and  SQ  are  the  intersections  of 
the  conical  surface  and  the  cutting  plane. 

.".  the  section  SPQ  is  a triangle,  Art.  81. 

( for  it  is  hounded  by  three  straight  lines). 

Q.  E.  D. 


Ex.  What  kina  of  triangle  is  a section  of  a right  circular  cone 
made  by  a plane  through  the  vertex  ? 


416 


BOOK  VIII.  SOLID  GEOMETRY 


Proposition  YII.  Theorem 

718.  Every  section  of  a circular  cone  made  by  a plane 
parallel  to  the  base  is  a circle. 


Given  the  circular  cone  SAB  with  apb  a section  made 
by  a plane  parallel  to  the  base. 

To  prove  apb  a circle. 

Proof.  Denote  the  center  of  the  base  by  0,  and  draw 
the  axis,  SO,  piercing  the  plane  of  the  section  in  o. 

Through  SO  and  any  element,  SP,  of  the  conical  sur- 
face, pass  a plane  cutting  the  plane  of  the  base  in  the 
radius  OP,  and  the  plane  of  the  section  in  op. 


In  like  manner,  pass  a plane  through  SO  and  SB  form- 
ing the  intersections  OB  and  ob. 


OP  ||  op,  and  OB  ||  ob. 

(Why  ?) 

A SPO 

and  SBO  are  similar  to 

A Spo  and  Sbo, 

respectively. 

Art.  328. 

op  °b 

OP~  \SO)  ~ OB 

(Why  ?) 

But 

OP=  OB. 

(Why?) 

.’.  op  = ob. 

(Wh7?) 

.*.  opb  is  a circle. 

(Why?) 

Q.  E.  D. 

719.  Cor.  The  axis  of  a circular  cone  passes  through 
the  center  of  every  section  parallel  to  the  base. 


CONES 


417 


Proposition  VIII.  Theorem 

720.  The  lateral  area  of  a cone  of  revolution  is  equal  to 
half  the  product  of  the  slant  height  by  the  circumference  of 
the  base. 


Given  a cone  of  revolution  having  its  lateral  area  de- 
noted by  8,  its  slant  height  by  L,  and  the  circumference 
of  its  base  by  C. 

To  prove  S=§  G X L. 

Proof.  Let  a regular  pyramid  be  circumscribed  about 
the  cone. 

Denote  the  lateral  area  of  the  pyramid  by  S',  and  the 
perimeter  of  its  base  by  P. 

Then  S'  = i P X L.  Art.  642. 

If  the  number  of  lateral  faces  of  the  circumscribed 
pyramid  be  indefinitely  increased, 

S'  will  approach  8 as  a limit.  Art.  715,  2.  , 
P will  approach  das  a limit.  Art.  441. 

And  \ P XL  will  approach  £ CXL  as  a limit.  Art.  253,  2. 

But  S'  — £ P X L always.  (Why?) 

:.8=iCXL.  (Why?) 

Q.  E.  D. 

721.  Formulas  for  lateral  area  and  total  area  of  a cone 
of  revolution.  Denoting  the  radius  of  the  base  by  E, 

S=£  (2  tiB  XL ) S=nRL. 

Also  T=nRLf-nRr  T=nB  (L+R), 


AA 


418 


BOOK  VIII. 


SOLID  GEOMETRY 


Proposition  IX. 

i 


Theorem 


722.  The  volume  of  a circular  cone  is  equal  to  one-third 
of  the  product  of  its  lase  by  its  altitude. 


Given  a circular  cone  having  its  volume  denoted  by  V, 
its  base  by  B,  and  its  altitude  by  H. 

To  prove  V—^BViH. 

Proof.  Let  a pyramid  with  a regular  polygon  for  its 
base  be  inscribed  in  the  given  cone. 

Denote  the  volume  of  the  inscribed  pyramid  by  V' , and 


its  base  by  B' . 

Hence  F = Xfl.  Art.  651. 

If  the  number  of  lateral  faces  of  the  inscribed  pyramid 
be  indefinitely  increased, 

V'  will  approach  Fas  a limit.  (Why  ?) 

B'  will  approach  B as  a limit.  (Why  ?) 

And  J B'  X H will  approach  £ B X H as  a limit.  (Why  ?) 
But  Y' — $ B'  X H always.  (Why?) 

.*.  V=h  B X E.  (Why  ?) 

Q.  E.  D. 

723.  Formula  for  the  volume  of  a circular  cone. 

F— & 7ii m. 


Ex.  1.  If,  in  a eone  of  revolution,  H=  3 and  B= 4,  find  S,  T and  V. 

Ex.  2.  If  the  altitude  of  a eone  of  revolution  equals  the  radius  of 
the  base,  what  do  the  formulas  for  S,  T and  V become  ? 


CONES 


419 


Proposition  X. 


Theorem 


724.  The  lateral  areas,  or  the  total  areas,  of  two  simi- 
lar cones  of  revolution  are  to  each  other  as  the  squares  of 
their  radii,  or  as  the  squares  of  their  altitudes,  or  as  the 
squares  of  their  slant  heights;  and  their  volumes  are  to  each 
other  as  the  cubes  of  these  lines. 


Given  two  similar  cones  of  revolution  having  their 
lateral  areas  denoted  by  S and  S',  their  total  areas  by  T 
and  T , their  volumes  by  V and  V , their  radii  by  R and 
R',  their  altitudes  by  H and  S',  and  their  slant  heights  by 
L and  L',  respectively. 

To  prove  S : S'=T  : T'  = R2:  R'2  = E2  : H'2  = I?  : L'2-, 


and  V : V'  = R3  : R'3  = E 3 : E'3  = L3  : L'3 . 

_ , ff  R L L + R 

Proof.  w R/  L,  L,  + Er 


(Why  ?) 


S = KRL  =R  L = R2  =L 2 = E2 
S'  nR'L'  E L'  R'2  L'2  E'2'  (Why?) 
T 7tR  (L  + R)  R ^ L + R _ R2  L2  E2 

T 7tR'(L'  + R')  R'^L'  + R'  R'2  L'2  E'2' 

(Why?) 

V §7tR2E  _R2  „E  _R3  _E3  _L3  „ 

F hnR'2E'  R'2  X W R'3  E'3  L'3'  (Why?) 

Q.  E.  D. 


725.  Dep.  An  equilateral  cone  is  a cone  of  revolution 
such  that  a section  through  the  axis  is  an  equilateral 
triangle. 


420 


BOOK  VIII.  SOLID  GEOMETRY 


Proposition  XI.  Theorem 

726.  The  lateral  area  of  a frustum  of  a cone  of  revolu- 
tion is  equal  to  one-lialf  the  sum  of  the  circumferences  of  its 
bases  multiplied  by  its  slant  height. 


Given  a frustum  of  a cone  of  revolution  having  its 
lateral  area  denoted  by  8,  its  slant  height  by  L,  the  radii 
of  its  bases  by  B and  r,  and  the  circumferences  of  its  bases 
by  C and  c. 

To  prove  S=i(C+c)XL. 

Proof.  Let  the  frustum  of  a regular  pyramid  be  circum- 
scribed about  the  given  frustum.  Denote  the  lateral  area 
of  the  circumscribed  frustum  by  S',  the  perimeter  of  the 
lower  base  by  P,  and  the  perimeter  of  the  upper  base  by  p. 

The  slant  height  of  the  circumscribed  frustum  is  L. 

Hence  S'  = b (P  + p)  XI.  Art.  643. 

Let  the  pupil  complete  the  proof. 

Q.  E.  D. 

721 . Formula  for  the  lateral  area  of  a frustum  of  a cone 
of  revolution.  8= J (2  nB  + 2 nr)  L. 

S=n  ( B + r ) L. 

728.  Cor.  The  lateral  area  of  a frustum  of  a cone  of 
revolution  is  equal  to  the  product  of  the  circumference  of  its 
midsection  by  its  slant  height. 


CONES 


421 


Proposition  XII.  Theorem 


729.  The  volume  of  the  frustum  of  a circular  cone  is 
equivalent  to  the  volume  of  three  cones,  whose  common  alti- 
tude is  the  altitude  of  the  frustum,  and  whose  bases  are  the 
lower  base,  the  upper  base,  and  a mean  proportional  between 
the  two  bases. 


Given  a frustum  of  a circular  cone  having  its  volume 
denoted  by  V,  its  altitude  by  H,  the  area  of  its  lower  base 
by  B,  and  that  of  its  upper  base  by  b. 

To  prove  T = ^ E {B  + b + V B X b) . 

Proof.  Let  the  frustum  of  a pyramid  with  regular  poly- 
gons for  its  bases  be  inscribed  in  the  given  frustum. 
Denote  the  volume  of  the  inscribed  frustum  by  V' , and  the 
areas  of  its  bases  by  B'  and  b' . 

.-.  V'  = h H (B'  + b'  + VB1  X b').  (Why?) 

If  the  number  of  lateral  faces  of  the  inscribed  frustum 
be  indefinitely  increased,  T7  will  approach  V,  B'  and  br 
approach  B and  b respectively,  and  Bf  X b'  approach  B X 
b,  as  limits.  Art.  715. 

Hence,  also,  B'  4*  b'-\-V  B'  X b'  will  approach  B+5  + 
Vb  Xb  as  a limit.  Art.  253. 

But  V — J H ( B'  -f  V +y/B’  X b')  always.  (Why  ?) 

F=$  H {B  + b + VBX  b).  (Why?) 

Q.  E.  D. 


422 


BOOK  Vin.  SOLID  GEOMETRY 


730.  Formula  for  the  volume  of  the  frustum  of  a circu- 
lar cone.  

V=  i H (nR2  + nr1  + X nr2) . 

/.  V=%  nH  (R2  + r2  + Rr) . 

Ex.  1.  The  measurement  of  the  volume  of  a frustum  of  a cone  of 
revolution  reduces  to  thej  measurement  of  the  lengths  of  what  straight 
lines? 

Ex.  2.  If  a conical  oil-can  is  12  in.  high,  how  much  more  tin  is 
required  to  make  it  than  to  make  a similar  oil-can  6 in.  high  ? How 
much  more  oil  will  it  hold  ? 

Ex.  3.  The  linear  dimensions  of  a conical  funnel  are  three  times 
those  of  a similar  funnel.  How  much  more  tin  is  required  to  make 
the  first  f How  much  more  liquid  will  it  hold  ? 

Ex.  4.  Make  a similar  comparison  of  cylindrical  oil-tanks.  Of 
conical  canvas  tents. 


EXERCISES.  CROUP  70 

THEOREMS  CONCERNING  CYLINDERS  AND  CONES 

Ex.  1.  Any  section  of  a cylinder  of  revolution  through  its  axis  is 
a rectangle. 

Ex.  2.  On  a cylindrical  surface  only  one  straight  line  can  be 
drawn  through  a given  point. 

[Sug.  For  if  two  straight  lines  could  be  drawn,  etc.] 

Ex.  3.  The  intersection  of  two  planes  tangent  to  a cone  is  a 
straight  line  through  the  vertex. 

Ex.  4.  If  two  planes  are  tangent  to  a cylinder,  their  line  of  inter- 
section is  parallel  to  an  element  of  the  cylinder. 

[Sug.  Pass  a plane  -L  to  the  elements  of  the  cylinder.] 

Ex.  5.  If  tangent  planes  be  passed  through  two  diametrically 
opposite  elements  of  a circular  cone,  these  planes  intersect  in  a 
straight  line  through  the  vertex  and  parallel  to  the  plane  of  the  base. 

Ex.  6.  In  a cylinder  of  revolution  the  diameter  of  whose  base 
equals  the  altitude,  the  volume  equals  one-third  the  product  of  the 
total  surface  by  the  radius  of  the  base* 


EXERCISES  ON  THE  CYLINDER  AND  CONE  423 


Ex.  7.  A cylinder  and  a cone  of  revolution  have  the  same  base 
and  the  same  altitude.  Find  the  ratio  of  their  lateral  surfaces,  and  also 
of  their  volumes. 

Ex.  8.  If  an  equilateral  triangle  whose  side  is  a be  revolved  about 
one  of  its  sides  as  an  axis,  find  the  area  generated  in  terms  of  a. 

Ex.  9.  If  a rectangle  whose  sides  are  a and  b be  revolved  first 
about  the  side  a as  an  axis,  and  then  about  the  side  6,  find  the  ratio  of 
the  lateral  areas  generated,  and  also  of  the  volumes. 

Ex.  10.  The  bases  of  a cylinder  and  of  a cone  of  revolution  are 
concentric.  The  two  solids  have  the  same  altitude,  and  the  diameter 
of  the  base  of  the  cone  is  twice  the  diameter  of  the  base  of  the  cylin- 
der. What  kind  of  line  is  the  intersection  of  their  lateral  surfaces, 
and  how  far  is  it  from  the  base  ? 

Ex.  11.  Determine  the  same  when  the  radius  of  the  cone  is  three 
times  the  radius  of  the  cylinder.  Also  when  r times. 

Ex.  12.  Obtain  a formula  in  terms  of  r for  the  volume  of  the 
frustum  of  an  equilateral  cone,  in  which  the  radius  of  the  upper  base 
is  r and  that  of  the  lower  base  is  3 r. 

Ex.  13.  A regular  hexagon  whose  side  is  a revolves  about  a diag- 
onal through  the  center  as  axis.  Find,  in  terms  of  a,  the  surface 
and  volume  generated. 

Ex.  14.  Find  the  locus  of  a point  at  a given  distance  from  a given 
straight  line. 

Ex.  15.  Find  the  locus  of  a point  whose  distance  from  a given 
line  is  in  a given  ratio  to  its  distance  from  a fixed  plane  perpen- 
dicular to  the  line. 

Ex  16.  Find  the  locus  of  all  straight  lines  which  make  a given 
angle  with  a given  line  at  a given  point. 

Ex.  17.  Find  the  locus  of  all  straight  lines  which  make  a given 
angle  with  a given  plane  at  a given  point. 

Ex.  18.  Find  the  locus  of  all  points  at  agiven  dis- 
tance from  the  surface  of  a given  cylinder  of  revolution. 

Ex.  19.  Find  the  locus  of  all  points  at  a given  dis- 
tance from  the  surface  of  a given  cone  of  revolution. 


424 


BOOK  Vm.  SOLID  GEOMETRY 


EXERCISES.  CROUP  71 

PROBLEMS  CONCERNING  THE  CYLINDER  AND  CONE 

Ex.  1.  Through  a given  element  of  a circular  cylinder,  pass  a 
plane  tangent  to  the  cylinder. 

Ex.  2.  Through  a given  element  of  a circular  cone,  pass  a plane  tangent 
to  the  cone. 

Ex.  3.  About  a given  circular  cylinder  circumscribe  a prism,  with 
a regular  polygon  for  its  base. 

Ex.  4.  Through  a given  point  outside  a circular  cylinder,  pass  a 
plane  tangent  to  the  cylinder. 

Ex.  5.  Through  a given  point  outside  a given  circular  cone,  pass 
a plane  tangent  to  the  cone. 

[Sug.  Through  the  vertex  of  the  cone  and  the  given  point  pass  a 
line,  and  produce  it  to  meet  the  plane  of  the  base.] 

Ex.  6.  Into  what  segments  must  the  altitude  of  a cone  of  revolution- 
be  divided  by  a plane  parallel  to  the  base,  in  order  that  the  volume  oi 
the  cone  be  bisected  ? 

Ex.  7.  Divide  the  lateral  surface  of  a given  cone  of  revolution  into 
two  equivalent  parts  by  a plane  parallel  to  the  base. 

Ex.  8.  If  the  lateral  surface  of  a cylinder  of  revolution  be  cut 
along  one  element  and  unrolled,  what  sort  of  a plane  figure  is  formed  ? 

Hence,  out  of  cardboard  construct  a cylinder  of  revolution  with 
given  altitude  and  given  circumference. 

Ex.  9.  If  the  lateral  surface  of  a cone  of  revolution  be  cut  along 
one  element  and  unrolled,  what  sort  of  a plane  figure  is  formed  1 

Hence,  out  of  cardboard  construct  a cone  of  revolution  of  given 
slant  height. 

Ex.  10.  Construct  an  equilateral  cone  out  of  pasteboard. 

Ex.  11.  Construct  a frustum  of  a cone  of  revolution  out  of  paste- 
board. 


i500K  IX 
THE  SPHERE 


731.  A sphere  is  a solid  bounded  by  a surface  all 
points  of  which  are  equally  distant  from  a point  within 
called  the  center. 


732.  A sphere  may  also  be  defined  as  a solid  generated 
by  the  revolution  of  a semicircle  about  its  diameter  as  an 
axis. 

Some  of  the  properties  of  a sphere  may  be  obtained  more  readily 
from  one  of  the  two  definitions  given,  and  some  from  the  other. 

A sphere  is  named  by  naming  the  point  at  its  center,  or  by  naming 
three  or  more  points  on  its  surface. 


733.  A radius  of  a sphere  is  a line  drawn  from  the 
center  to  any  point  on  the  surface. 

A diameter  of  a sphere  is  a line  drawn  through  the 
center  and  terminated  at  each  end  by  the  surface  of  the 
sphere. 

734.  A line  tangent  to  a sphere  is  a line  having  but  one 
point  in  common  with  the  surface  of  the  sphere,  however 
far  the  line  be  produced. 


(425) 


426 


BOOK  IX.  SOLID  GEOMETRY 


735.  A plane  tangent  to  a sphere  is  a plane  having  but 
one  point  in  common  with  the  surface  of  the  sphere,  how- 
ever far  the  plane  be  produced. 

736.  Two  spheres  tangent  to  each  other  are  spheres 
whose  surfaces  have  one  point,  and  only  one,  in  common. 

737.  Properties  of  a sphere  inferred  immediately. 

1.  All  radii  of  a sphere,  or  of  equal  spheres,  are  equal. 

2.  All  diameters  of  a sphere,  or  of  equal  spheres,  are 
equal. 

3.  Two  spheres  are  equal  if  their  radii  or  their  diame- 
ters are  equal. 


Proposition  I.  Theorem 

738.  A section  of  a sphere  made  by  a plane  is  a circle. 


Given  the  sphere  0,  and  PCD  a section  made  by  a plane 
cutting  the  sphere. 

To  prove  that  PCD  is  a circle. 

Proof.  From  the  center  0,  draw  OA  _L  the  plane  of 
the  section. 


THE  SPHERE 


427 


Let  C be  a fixed  point  on  the  perimeter  of  the  section, 


and  P any  other  point  on  this  perimeter. 

Draw  AC,  AP,  OC,  OP. 

Then  the  A OAP  and  OAC  are  rt.  A.  Art.  505. 

OP=OC.  (Why?) 

OA  = OA.  (Why?) 

A OAP=  A OAC.  (Why?) 

AP=AC.  (Why?) 

But  P is  any  point  on  the  perimeter  of  the  section  PCD. 
:.  every  point  on  this  perimeter  is  at  the  distance  AC 
from  A. 

PCD  is  a circle  with  center  A.  Art.  197. 

Q.  E.  D. 


739.  Cor.  1.  Circles  ivl rich  are  sections  of  a sphere 
made  by  planes  equidistant  from  the  center  are  equal;  and 
conversely . 

740.  Cor.  2.  Of  two  circles  on  a sphere,  the  one  made 
by  a plane  more  remote  from  the  center  is  smaller;  and 
conversely. 

741.  Def.  A great  circle  of  a sphere  is  a circle  whose 
plane  passes  through  the  center  of  the  sphere. 

742.  Def.  A small  circle  of  a sphere  is  a circle  whose 
plane  does  not  pass  through  the  center  of  the  sphere. 

743.  Def.  The  axis  of  a circle  of  a sphere  is  the 

diameter  of  a sphere  which  is  perpendicular  to  the  plane  of 
the  circle.  Thus,  on  figure  p.  426,  BB'  is  the  axis  of  PCD. 

744.  Def.  The  poles  of  a circle  of  a sphere  are  the 

extremities  of  the  axis  of  the  circle.  Thus,  B and  B' , of 
figure  p.  426,  are  poles  of  the  circle  PCD. 


428 


BOOK  IX.  SOLID  GEOMETRY 


745.  Properties  of  circles  of  a sphere  inferred  imme- 
diately. 

1.  The  axis  of  a circle  of  a sphere  passes  through  the 
center  of  the  circle;  and  conversely. 

2.  Parallel  circles  have  the  same  axis  and  the  same  poles. 

3.  All  great  circles  of  a sphere  are  equal. 

4.  Every  great  circle  on  a sphere  bisects  the  sphere  and 
its  surface. 

5.  Two  great  circles  on  a sphere  bisect  each  other. 

For  the  line  of  intei’seetion  of  the  two  planes  of  the 
circles  passes  through  the  center,  and  hence  is  a diameter 
of  each  circle. 

6.  Through  tic o points  {not  the  extremities  of  a diameter) 
on  the  surface  of  a sphere,  one,  and  only  one,  great  circle  can 
be  passed. 

For  the  plane  of  the  great  circle  must  also  pass  through 
the  center  of  the  sphere  (Art.  741),  and  through  three 
points  not  in  a straight  line  only  one  plane  can  be  passed 
(Art.  500). 

7.  Through  any  three  points  on  the  surface  of  a sphere, 
not  in  the  same  plane  with  the  center,  one  small  circle,  and 
only  one,  can  be  passed. 

746.  Def.  The  distance  between  two  points  on  the  sur- 
face of  a sphere  is  the  length  of  the  minor  arc  of  a great 
circle  joining  the  points. 

Ex.  1.  If  the  radius  of  a sphere  is  13  in.,  find  the  radius  of  a 
circle  on  the  sphere  made  by  a plane  at  a distance  of  1 ft.  from  the 
center. 

Ex.  2.  What  geographical  circles  on  the  earth’s  surface  are  great, 
and  what  small  circles  ? 

Ex.  3.  What  is  the  largest  number  of  points  in  which  two  circles 
on  the  surface  of  a sphere  can  intersect  ? Why  ? 


THE  SPHERE 


429 


Proposition  II.  Theorem 

747.  All  points  in  the  circumference  of  a circle  of  a 
sphere  are  equally  distant  from  each  pole  of  the  circle. 


p 
A 


Given  ABC  a circle  of  a sphere,  and  P and  P'  its  poles. 

To  prove  the  arcs  PA,  PB,  PC  equal,  and  arcs  P'A, 
P'B,  P'C  equal. 

Proof.  Draw  the  chords  PA,  PB,  PC. 

The  chords  PA,  PB  and  PC  are  equal.  Art.  518. 

/.  arcs  PA,  PB  and  PC  are  equal.  Art.  218, 

In  like  manner,  the  arcs  P'A,  P'B  and  P'C  may  be 
proved  equal. 

Q.  E.  D. 

748.  Def.  The  polar  distance  of  a small  circle  on  a 
sphere  is  the  distance  of  any  point  on  the  circumference  of 
the  circle  from  the  nearer  pole. 

The  polar  distance  of  a great  circle  on  a sphere  is  the  dis- 
tance of  any  point  on  the  circumference  of  the  great  circle 
from  either  pole. 

749.  Cor.  The  polar  distance  of  a great  circle  is  the 
quadrant  of  a great  circle. 


430 


BOOK  IX.  SOLID  GEOMETEY 


Proposition  III.  Theorem 

750.  If  a point  on  the  surface  of  a sphere  is  at  a quad- 
rant’s distance  from  two  other  points  on  the  surface , it  is 
the  pole  of  the  great  circle  through  those  points. 


Given  PB  and  PC  quadrants  on  the  surface  of  the 
sphere  0,  and  ABC  a great  circle  through  B and  C. 

To  prove  that  P is  the  pole  of  ABC. 

Proof.  From  the  center  0 draw  the  radii  OB,  OC,  OP. 


The  arcs  PB  and  PC  are  quadrants.  (Why  ?) 

o'.  A POB  and  POC  are  rt.  A . (Why  ?) 

.'.  PO  -L  plane  ABC.  (Why?) 

.'.  P is  the  pole  of  the  great  circle  ABC.  (Why?) 


Q.  E.  D. 

751.  Cor.  Through  two  given  points  on  the  surface  of 
a sphere  to  describe  a great  circle. 

Let  A and  B be  the  given  points. 

From  A and  B as  centers,  with  a 
quadrant  as  radius,  describe  arcs 
on  the  surface  of  the  sphere  inter- 
secting at  P.  With  P as  a center 
and  a quadrant  as  a radius,  describe  a great  circle. 


THE  SPHERE 


431 


Proposition  IV.  Theorem 

752.  A plane  perpendicular  to  a radius  at  its  extremity 
is  tangent  to  the  sphere. 

c 


Given  the  sphere  0,  and  the  plane  MN  -L  the  radius  OA 
of  the  sphere  at  its  extremity  A. 

To  prove  MN  tangent  to  the  sphere. 

Proof.  Take  P any  point  in  plane  MN  except  A.  Draw 
\0P.  Then  OP  > OA.  (Why?) 

• \ the  point  P is  outside  the  surface  of  the  sphere. 

But  P is  any  point  in  the  plane  MN  except  A. 

. \ plane  AIN  is  tangent  to  the  sphere  at  the  point  A, 

( for  every  point  in  the  plane,  except  A,  is  outside  the  surface  of  the  sphere). 

Art.  735.  Q.  E.  D. 

753.  Cor.  1.  A plane,  or  a line,  which'  is  tangent  to  a 
sphere,  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact.  Also,  if  a plane  is  tangent  to  a sphere,  a perpendicu- 
lar to  the  plane  at  its  point  of  contact  passes  through  the  center 
of  the  sphere. 

754.  Cor.  2.  A straight  line  perpendicular  to  a radius 
of  a sphere  at  its  extremity  is  tangent  to  the  sphere. 

755.  Cor.  3.  A straight  line  tangent  to  a circle  of  a 
sphere  lies  in  the  plane  tangent  to  the  sphere  at  the  point  of 
contact, 


432 


BOOK  IX.  SOLID  GEOMETRY 


756.  Cor.  4.  A straight  line  drawn  in  a tangent  plane, 
and  through  the  point  of  contact  is  tangent  to  the  spjhere  at 
that  point. 

757.  Cor.  5.  Two  straight  lines  tangent  to  a sphere  at 
a given  point  determine  the  tangent  plane  at  that  point. 

758.  Def.  A sphere  circumscribed  about  a polyhedron. 

is  a sphere  in  whose  surface  lie  all  the  vertices  of  the 
polyhedron. 

759.  Def.  A sphere  inscribed  in  a polyhedron  is  a 
sphere  to  which  all  the  faces  of  the  polyhedron  are  tangent. 


Proposition  V.  Problem 

760.  To  circumscribe  a sphere  about  a given  tetra- 
hedron. 


D 


Given  the  tetrahedron  ABCD. 

To  circumscribe  a sphere  about  ABCD. 

Construction  and  Proof.  Construct  E and  F the  centers 
of  circles  circumscribed  about  the  A ABC  and  BCD , re- 
spectively. Art.  286. 

Draw  EH  _L  plane  ABC  and  FK  JL  plane  BCD.  Art.  514. 
Draw  EG  and  FG  to  G the  midpoint  of  BC. 


THE  SPHERE 

433 

EG  and  FG  are  A BG. 

Art.  113. 

:.  plane  EGF  A BG. 

Art.  509. 

:.  plane  EGF  JL  plane  ABG. 

Art.  555. 

:.  EE  lies  in  the  plane  FGE. 

Art.  558. 

In  like  manner  FK  lies  in  tlie  plane  FOE. 

The  lines  EG  and  FO  are  not  ||, 

( for  they  meet  in  the  point  G-). 

:.  the  lines  EH  and  FK  are  not  j|.  Art.  122. 

Hence  EE  must  meet  FK  in  some  point  0. 

But  EH  is  the  locus  of  all  points  equidistant  from  A, 
B and  C\  and  FK  is  the  locus  of  all  points  equidistant 
from  B,  C and  F).  Art.  520. 

0,  which  is  in  both  EE  and  FK,  is  equidistant  from 
A,  B,  C and  D.  (Why?) 

Hence  a spherical  surface  constructed  with  0 as  a center 
and  OA  as  a radius  will  pass  through  A,  B,  C and  E,  and 
form  the  sphere  required.  q.  e.  f. 

761.  Cor.  1.  Four  points  not  in  the  same  plane  deter- 
mine a sphere. 

762.  Cor.  2.  The  four  perpendiculars  erected  at  the 
centers  of  the  faces  of  a tetrahedron  meet  in  a point. 

763.  Cor.  3.  The  six  planes  perpendicular  to  the  edges 
of  a tetrahedron  at  their  midpoints  intersect  in  a point. 

764.  Def.  An  angle  formed  by  two  curves  is  the  angle 
formed  by  a tangent  to  each  curve  at  the  point  of  inter- 
section. 

765.  Def.  A spherical  angle  is  an  angle  formed  by 
two  intei’secting  arcs  of  great  circles  on  a sphere,  and 
hence  by  tangents  to  these  arcs  at  the  point  of  intersection. 

BB 


434 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  VI.  Problem 


766.  To  inscribe  a sphere  in  a given  tetrahedron . 


Given  the  tetrahedron  ABCD. 

To  inscribe  a sphere  in  ABCD. 

Construction  and  Proof.  Bisect  the  dihedral  angle  D- 
AB~ C by  the  plane  OAB;  similarly  bisect  the  dihedral  A 
whose  edges  are  BC  and  AC  by  the  planes  OBC  and  OAC, 
respectively. 

Denote  the  point  in  which  the  three  bisecting  planes 
intersect  by  0. 

Every  point  in  the  plane  OAB  is  equidistant  from  the 
faces  DAB  and  CAB.  Art.  562. 

Similarly,  every  point  in  OBC  is  equidistant  from  the 
two  faces  intersecting  in  BC,  and  every  point  in  OAC  is 
equidistant  from  the  two  faces  intersecting  in  AC. 

0 is  equidistant  from  all  four  faces  of  the  tetra- 
hedron. Ax.  1. 

Hence,  from  0 as  a center,  with  the  J_  from  0 to  any 
one  face  as  a radius,  describe  a sphere. 

This  sphere  will  be  tangent  to  the  four  faces  of  the 

tetrahedron  and  inscribed  in  the  tetrahedron.  Art.  759. 

Q.  E.  F. 

767.  Cor.  The  planes  bisecting  the  six  dihedral  angles 
of  a tetrahedron  meet  in  one  point. 


THE  SPHERE 


■435 


Proposition  VII.  Problem 
768.  To  find  the  radius  of  a given  material  sphere. 


Given  the  material  sphere  0. 

To  construct  the  radius  of  the  sphere. 

Construction.  With  any  point  P (Fig.  1)  of  the  surface 
of  the  sphere  as  a pole,  describe  any  convenient  circum- 
ference on  the  surface. 

On  this  circumference  take  any  three  points  A,  B and  G. 

Construct  the  A ABC  (Fig.  2)  having  as  sides  the  three 
chords  AB,  BC,  AG,  obtained  from  Fig.  1,  by  use  of  the 
compasses.  Art.  283. 

Circumscribe  a circle  about  the  A ABG.  Art.  286. 

Let  KB  be  the  radius  of  this  circle. 

Construct  (Fig.  3)  the  right  A kpb,  having  for  hypot* 
enuse  the  chord  pb  (Fig.  1)  and  the  base  kb.  Art.  284. 

Draw  bp ' _L  bp  and  meeting  pk  produced  at  p' . 

Bisect^'  at  0. 

Then  op  is  the  radius  of  the  given  sphere. 

Proof.  Let  the  pupil  supply  the  proof, 


Q.  E.  F. 


436 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  VIII.  Theorem 

769.  The  intersection  of  two  spherical  surfaces  is  the 
circumference  of  a circle  whose  plane  is  perpendicular  to  the 
line  joining  the  centers  of  the  spheres,  and  whose  center  is  in 
that  line. 


Given  two  intersecting  ® 0 and  O'  which,  by  rotation 
about  the  line  00'  as  an  axis,  generate  two  intersecting 
spherical  surfaces. 

To  prove  that  the  intersection  of  the  spherical  surfaces 
is  a O,  whose  plane  _L  00' , and  whose  center  lies  in  00'. 

Proof.  Let  the  two  circles  intersect  in  the  points  P and 
Q,  and  draw  the  common  chord  PQ. 

Then,  as  the  two  given  ® rotate  about  00'  as  an  axis, 
the  point  P will  generate  the  line  of  intersection  of  the  two 
spherical  surfaces  that  are  formed. 

But  PR  is  constantly  J_  00' . Art.  241. 

/.  PR  generates  a plane  L OO1  Art.  510. 

Also  PR  remains  constant  in  length. 

.•.  P describes  a circumference  in  that  plane.  Art.  197. 

Hence  the  intersection  of  two  spherical  surfaces  is  a O, 
whose  plane  _L  the  line  of  centers,  and  whose  center  is  in 
the  line  of  centers. 

Q.  E.  D. 

The  above  demonstration  is  an  illustration  of  the  use  of  the  second 
definition  of  a sphere  (Art.  732). 


THE  SPHERE 


437 


Proposition  IX.  Theorem 

770.  A spherical  angle  is  measured  by  the  arc  of  a great 
circle  described  from  the  vertex  of  the  angle  as  a pole , and 
included  between  its  sides , produced , if  necessary. 


Given  ABAC  a spherical  angle  formed  by  the  intersec- 
tion of  the  arcs  of  the  great  circles  BA  and  CA,  and  BC 
an  arc  of  a great  circle  whose  pole  is  A. 

To  prove  ABAC  measured  by  arc  BC. 

Proof.  Draw  AT)  tangent  to  AB,  and  AF  tangent  to 
AC.  Also  draw  the  radii  OB  and  OC. 


Then 

AB  A.  AO. 

Art.  230. 

Also 

OB  A AO  {for  AB  is  a quadrant). 

OB  \\AD. 

(Why?) 

Similarly 

OC  ||  AF. 

A BOC=  A DAF 

Art.  538. 

But  A BOG  is 

measured  by  arc  BC. 

Art.  257. 

.’.  ABAF,  that  is,  ABAC,  is  measured  by 

arc  BC 

(Why?) 

Q.  E.  D. 

771.  Cor.  A spherical  angle  is  equal  to  the  plane  angle 
of  the  dihedral  angle  formed  by  the  planes  of  its  sides, 


438 


BOOK  IX.  SOLID  GEOMETEY 


SPHERICAL  TRIANGLES  AND  POLYGONS 

772.  A spherical  polygon  is  a portion  of  the  surface  of 
a sphere  bounded  by  three  or  more 
arcs  of  great  circles,  as  ABCD. 

The  sides  of  the  spherical  polygon 
are  the  bounding  arcs;  the  vertices 
are  the  points  in  which  the  sides  in- 
tersect; the  angles  are  the  spherical 
angles  formed  by  the  sides. 

The  sides  of  a spherical  polygon  are  usually  limited  to  arcs  less 
than  a semicircumference. 

773.  A spherical  triangle  is  a spherical  polygon  of  three 
sides. 

Spherical  triangles  are  classified  in  the  same  way  as 
plaue  triangles;  viz.,  as  isosceles,  equilateral,  scalene, 
right,  obtuse  and  acute. 

774.  Relation  of  spherical  polygons  to  polyhedral  angles. 

If  radii  be  drawn  from  the  center  of  a sphere  to  the  ver- 
tices of  a spherical  polygon  on  its  surface  (as  OJL,  OB, 
etc.,  in  the  above  figure),  a polyhedral  angle  is  formed  at 
0,  which  has  an  important  relation  to  the  spherical  poly- 
gon ABCD 

Each  face  angle  of  the  'polyhedral  angle  equals  ( in  num- 
ber of  degrees  contained)  the  corresponding  side  of  the  spheri- 
cal polygon; 

Each  dihedral  angle  of  the  polyhedral  angle  equals  the 
corresponding  angle  of  the  spherical  polygon. 

Hence,  corresponding  to  each  property  of  a polyhedral 
angle,  there  exists  a property  of  a spherical  polygon , and 
conversely. 


THE  SPHERE 


439 


Hence,  also,  a trihedral  angle  and  its  parts  correspond 
to  a spherical  triangle  and  its  parts. 

Of  the  common  properties  of  a polyhedral  angle  and  a spherical 
polygon,  some  are  discovered  more  readily  from  the  one  figure  and 
some  from  the  other.  In  general,  the  spherical  polygon  is  simpler 
to  deal  with  than  a polyhedral  angle.  For  instance,  if  a trihedral 
angle  were  drawn  with  the  plane  angles  of  its  dihedral  angles,  nine 
lines  would  be  used,  forming  a complicated  figure  in  solid  space ; 
whereas,  the  same  magnitudes  are  represented  in  a spherical  triangle 
by  three  lines  in  an  approximately  plane  figure. 

On  the  other  hand,  the  spherical  polygon,  because  of  its  lack  of 
detailed  parts,  is  often  not  so  suggestive  of  properties  as  the  poly- 
hedral angle. 


Proposition  X.  Theorem 

775.  The  sum  of  two  sides  of  a spherical  triangle  is 
greater  than  the  third  side. 


Given  the-  spherical  triangle  ABC,  of  which  no  side  is 
larger  than  AB. 

To  prove  AC  + BC  > AB. 

Proof.  From  the  center  of  the  sphere,  0,  draw  the  radii 
OA,  OB,  OC. 

Then,  in  the  trihedral  angle  O-ABC, 

ZAOC-f  Z BOG  > AOB.  Art.  582. 

AC  + BC  > AB.  Art.  774. 

q.  e.  d. 


440 


BOOK  IX.  SOLID  GEOMETRY 


776.  Cor.  1.  Any  side  of  a spherical  triangle  is  greater 
than  the  difference  between  the  other  two  sides. 

Ill . Cor.  2.  The  shortest  path  bettveen  two  points  on 
the  surface  of  a sphere  is  the  arc  of  the  great  circle  joining 
those  points. 

For  any  other  path  between  the  two  points  may  be 
made  the  limit  of  a series  of  arcs  of  great  circles  connect- 
ing successive  points  on  the  path,  and  the  sum  of  this 
series  of  arcs  of  great  circles  connecting  the  two  points  is 
greater  than  the  single  arc  of  a great  circle  connecting 
them. 


Proposition  XI.  Theorem 

778.  The  sum  of  the  sides  of  a spherical  polygon  is  less 
than  360°. 


Given  the  spherical  polygon  ABGD. 

To  prove  the  sum  of  the  sides  of  ABCB  < 360°. 

Proof.  From  0,  the  center  of  the  sphere,  draw  the  radii 
OA , OB,  OC,  OB. 

Then  AAOBA  ABOC+  ACOB  + IDOA  < 3G0°. 

(Why?) 

AB  ABC  A CD  A BA  < 360°.  Art.  774. 

Q.  E.  D. 


SPHERICAL  TRIANGLES 


441 


779.  Def.  The  polar 
triangle  of  a given  triangle 
is  the  triangle  formed  by 
taking  the  vertices  of  the 
given  triangle  as  poles,  and 
describing  arcs  of  great  cir- 
cles. (Hence,  if  each  pole 
be  regarded  as  a center,  the  radius  used  in  describing  each 
arc  is  a quadrant.)  Thus  A'B'G'  is  the  polar  triangle  of 
ABC ; also  D'E'F'  is  the  polar  triangle  of  DEF. 


Proposition  XII.  Theorem 

780.  If  one  spherical  triangle  is  the  polar  of  another, 
then  the  second  triangle  is  the  polar  of  the  first. 

a' 


Given  A'B'C'  the  polar  triangle  of  ABC. 

To  prove  ABC  the  polar  triangle  of  A'B’C' . 


Proof.  B is  the  pole  of  the  arc  A'C' . Art.  779. 

.’.  arc  A'B  is  a quadrant.  (Why  ?) 

Also  C is  the  pole  of  the  arc  A'B'.  (Why  ?) 

/.  arc  A'C  is  a quadrant.  (Why?) 

.*.  A'  is  at  a quadrant’s  distance  from  both  B and  C. 

:.  A'  is  the  pole  of  the  arc  BC.  Art.  7-50. 

In  like  manner  it  may  be  shown  that  B'  is  the  pole  of 
AC,  and  C'  the  pole  of  AB.  q.  e.  d. 


442 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  XIII.  Theorem 

781.  In  a spherical  triangle  and  its  polar , each  angle  of 
one  triangle  is  the  supplement  of  the  side  opposite  in  the 
other  triangle. 

A' 

b' 


Given  the  polar  A ABC  and  A'B'C'  with  the  sides  of 
ABC  denoted  by  a,  b,  c,  and  the  sides  of  A'B'C'  denoted 
by  a',  b',  c',  respectively. 

To  prove  A + a'  = 180°,  B + b'=  180°,  C +c'=  180°r 
A'  + a = 180°,  B'+b=  180°,  C'  + c = 180°. 
Proof.  Produce  the  sides  AB  and  AC  till  they  meet 


B’C  in  the  points  B and  F,  respectively. 

Then  B'  is  the  pole  of  AF  arc  B'F=  90°.  Art.  7so. 

Also  C'  is  the  pole  of  AD  arc  C'D  = cJ0°.  (Why?) 

Adding,  B'F  JrC'D  = lS0° . (Why?) 

Or  B'F  + FC  + DF=  180° . Ax.  6. 

Or  B'C'  + DF=  180°. 


But  B'C'  = af,  and  DF  is  the  measure  of  the  Z A.  Art.  770. 

.'.  A + a'  -180°. 

In  like  manner  the  other  supplemental  relations  may  be 
proved  as  specified. 

Q.  E.  D. 

782.  Def.  Supplemental  triangles  are  two  spherical 
triangles  each  of  which  is  the  polar  triangle  of  the  other. 

This  new  name  for  two  polar  triangles  is  due  to  the 
property  proved  in  Art.  781, 


SPHERICAL  TRIANGLES 


443 


Proposition  XIV.  Theorem 

783.  The  sum  of  the  angles  of  a spherical  triangle  is 
greater  than  180°,  and  less  than  540°. 

jt 


Given  the  spherical  triangle  ABC. 

To  prove  A B C > 180°  and  < 540°. 

Proof.  Draw  A'B'C',  the  polar  triangle  of  ABC,  and 
denote  its  sides  by  a',  A,  c' . 


Then  A + a'  = 180°  ) 

B+V  = 180°  Y 

Art.  781. 

C + e'  = 1S0°  ) 

, A B C -j-  a'  -f-  V c'  — 540  . 

(1)  Ax.  2. 

But  ( a'  -f-  h'  -f-  o'  < 360 

Art.  778. 

| a'  + V + d > 0° 

Subtracting  each  of  these  in  turn  from  (1), 

A + B + O > 180°  and  < 540°. 

Ax.  11. 

Q.  E.  D. 


784.  Cor.  A spherical  triangle  may  have  one , two  or 
three  right  angles;  or  it  may  have  one,  two  or  three  obtuse 
angles. 

785.  Def.  A birectangular  spherical  triangle  is  a spher- 
ical triangle  containing  two  right  angles. 

786.  Def.  A trirectangular  spherical  triangle  is  a 
spherical  triangle  containing  three  right  angles. 


444 


BOOK  IX.  SOLID  GEOMETRY 


787.  Cor.  The  surface  of  a sphere  may  he  divided  into 
eight  trirectangular  spherical  triangles.  For  let  three  planes 
_L  to  each  other  be  passed  through  the  center  of  a sphere,  etc. 

788.  Def.  The  spherical  excess  of  a spherical  triangle 
is  the  excess  of  the  sum  of  its  angles  over  180°. 

789.  Def.  Symmetrical  spherical  triangles  are  triangles 
which  have  their  parts  equal,  but  arranged  in  reverse  order. 


Three  planes  passing  through  the  center  of  a sphere 
form  a pair  of  symmetrical  spherical  triaugles  on  opposite 
sides  of  the  sphere  (see  Art.  580),  as  A ABC  and  A'B'C' 
of  Fig.  1. 

790.  Equivalence  of  symmetrical  spherical  triangles. 
Two  plane  triangles  which  have  their  parts  equal,  but  ar- 
ranged inreverse  Q> 


triangle,  turning 

it  over  in  space,  and  placing  it  upon  the  other  triangle. 

But  two  symmetrical  spherical  triangles  cannot  be  made 
to  coincide  in  this  way,  because  of  the  curvature  of  a 
spherical  surface.  Hence  the  equivalence  of  two  sym- 
metrical spherical  triangles  must  be  demonstrated  in  some 
indirect  way. 


order,  may  be 
made  to  coincide 
by  lifting  up  one 


■A 


SPHERICAL  TRIANGLES 


445 


791.  Property  of  symmetrical  spherical  triangles.  Two 

isosceles  symmetrical  spherical  triangles  are  equal,  for  they 
can  be  made  to  coincide. 


Proposition  XV.  Theorem 
792.  Two  symmetrical  spherical  triangles  are  equivalent. 


Given  the  symmetrical  spherical  A ABC  and  A'B'C', 
formed  by  planes  passing  through  0,  the  center  of  a sphere. 
(See  Art.  789.) 

To  prove  a ABCoa  A'B'C'. 

Proof.  Let  P be  the  pole  of  a small  circle  passing  through 
the  points  A,  B,  C.  Draw  the  diameter  POP'. 

Also  draw  PA,  PB,  PC,  P'A',  P'B',  P'C',  all  arcs  of 
great  ©. 

Then  P.4  = PB  = PC.  Art.  747. 

Also  P'A'  = PA,  P'B'  = PB,  P'C'  = PC. 

Arts.  78,  215. 

.%  P'A'  = P'B'  =P'C'.  Ax.  1. 

Hence  PAB  and  P'A'B'  are  symmetrical  isosceles  A. 


.*.  A PAB  — A P'A'B' . 1 

Similarly 

A PAC=  A P'A'C'.  r 

Art.  791. 

And 

A PBG—  A P’B'C.) 

Adding 

A PAB  A-  A PA  0+  A PBG 

oAFiT'+A  P'A'C' A-  A P'B'G'. 

Ax.  2. 

Or 

A ABC^A  A'B'C'. 

Ax.  6. 

In  case  the  poles  P and  P'  fall  outside  the  A 

ABC  and 

A'B'C',  let  the  puoil  supply  the  demonstration.  q.  e.  d. 


446 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  XVI.  Theorem 

793.  On  the  same  sphere,  or  on  equal  spheres,  two  tri- 
angles are  equal, 

I.  If  two  sides  and  the  included  angle  of  one  are  equal  to 
two  sides  and  the  included  angle  of  the  other ; or 

II.  If  two  angles  and  the  included  side  of  one  are  equal 
to  two  angles  and  the  included  side  of  the  other, 

the  corresponding  equal  parts  being  arranged  in  the  same 
order  in  each  case. 


A d 


I.  Given  the  spherical  A ABC  and  DEF,  in  -which 
AC=DF,  CB  = FE,  and  ZC=ZF. 

To  prove  A ABC=  A DEF. 

Proof.  Let  the  pupil  supply  the  proof  (see  Book  I, 
Prop.  VI). 

II.  Given  the  spherical  A ABC  and  DEF,  in  -which 
ZC=ZF,  Z.B  — ZE,  and  CB  = FE. 

To  prove  A ABC=  A DEF. 

Proof.  Let  the  pupil  supply  the  proof  (see  Book  I, 
Prop.  VII) . 

Ex.  1.  If  the  line  of  centers  of  two  spheres  is  10  in.,  and  the  radii 
are  12  in.  and  3 in.,  how  are  the  spheres  situated  with  reference  to 
each  other  ? 

Ex.  2.  The  tank  on  a motor  car  is  a cylinder  35  inches  long  and 
15  inches  in  diameter.  How  many  gallons  of  gasolene  will  it  hold  ? 

Ex.  3.  In  an  equilateral  cone,  find  the  ratio  of  the  lateral  area  to 
the  area  of  the  base. 


SPHERICAL  TRIANGLES 


447 


Proposition  XVII.  Theorem 

794.  On  the  same  sphere , or  on  equal  spheres,  tivo  tri- 
angles are  symmetrical  and  equivalent , 

I.  If  tivo  sides  and  the  included  angle  of  one  are  equal  to 
two  sides  and  the  included  angle  of  the  other;  or 

II.  If  two  angles  and  the  included  side  of  one  are  equal 
to  two  angles  and  the  included  side  of  the  other, 

the  corresponding  equal  giants  being  arranged  in  reverse 
order. 


I.  Given  the  spherical  A ABC  and  DBF,  in  which  AB 
— BE,  AC  = DF,  and  /.A  — Z.D , the  corresponding  parts 
oeing  arranged  in  reverse  order. 

To  prove  A ABC  symmetrical  with  A DEF. 

Proof.  Construct  the  A D'E'F'  symmetrical  with  A DEF. 

Then  A ABC  may  be  made  to  coincide  with  A D'E'F' , 

Art.  793. 

( having  two  sides  and  the  included  Z equal  and  arranged  in  the 
same  order). 

But  A D'E'F'  is  symmetrical  with  the  A DEF. 

:.  A ABC,  which  coincides  with  A D'E'F' , is  symmet- 
rical with  A DEF. 

II.  The  second  part  of  the  theorem  is  proved  in  the 
same  way. 


Q.  £.  D. 


448 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  XVIII.  Theorem 


795.  If  two  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equilateral,  they  are  also  mutually 
equiangular , and  therefore  equal  or  symmetrical . 


Given  two  mutually  equilateral  spherical  A ABC  and 
A'B'C  on  the  same  or  on  equal  spheres. 

To  prove  A ABC  and  A'B'C'  equal  or  symmetrical. 

Proof.  From  0 and  O',  the  centers  of  the  spheres  to 
■which  the  given  triangles  belong,  draw  the  radii  OA,  OB, 
OC,  O’ A',  O'B',  O’C. 

Then  the  face  A at  0= corresponding  face  A at  O'. 

Art.  774. 

Hence  dihedral  A at  0 = corresponding  dihedral  A at  O'. 

Art.  584. 

.\  A of  spherical  A ABC— homologous  A of  spherical 
A A'B'C'.  Art.  774. 

.'.  the  A ABC  and  A'B'C'  are  equal  or  symmetrical, 
according  as  their  homologous  parts  are  arranged  in  the 
same  or  in  reverse  order.  Art.  789. 


796.  Note'.  The  conditions  in  Props.  XVT  and  XVIII  which  make 
two  spherical  triangles  equal  are  the  same  as  those  which  make 
two  plane  triangles  equal.  Hence  many  other  propositions  occur  in 
spherical  geometry  which  are  identical  with  corresponding  proposi- 
tions in  plane  geometry.  Thus,  many  of  the  construction  problems  of 
spherical  geometry  are  solved  in  the  same  way  as  the  corresponding 
construction  problems  in  plane  geometry;  as,  to  bisect  a given 
angle,  etc, 


SPHERICAL  TRIANGLES 


449 


Proposition  XIX.  Theorem 

797.  If  two  triangles  on  the  same  sphere  are  mutually 
equiangular , they  are  also  mutually  equilateral,  and  there- 
fore equal  or  symmetrical . 


Given  the  mutually  equiangular  spherical  A Q and  O' 
on  the  same  sphere  or  on  equal  spheres. 

To  prove  that  Q and  Q'  are  mutually  equilateral,  and 
therefore  equal  or  symmetrical. 

Proof.  Construct  P and  P'  the  polar  A of  Q and  O', 


respectively. 

Then  A P and  P'  are  mutually  equilateral.  Art.  781. 

/.  A P and  P'  are  mutually  equiangular.  Art.  795. 

But  Q is  the  polar  A of  P,  and  Q'  of  P' . Art.  780. 

/.  A Q and  Q'  are  mutually  equilateral.  Art.  781. 

Hence  Q and  Q'  are  equal  or  symmetrical,  according  as 
their  homologous  parts  are  arranged  in  the  same  or  in 
reverse  order.  Art.  789. 

Q.  E.  D. 


798.  Cor.  If  two  mutually  equiangular  triangles  are 
on  unequal  spheres , their  corresponding  sides  have  the  same 
ratio  as  the  radii  of  their  respective  spheres. 

cc 


450 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  XX.  Theorem 
799.  In  an  isosceles  spherical  triangle  the  angles  oppo* 


Given  the  spherical  A ABC  in  which  AB=AC. 

To  prove  ZB=  ZC. 

Proof.  Draw  an  arc  from  the  vertex  A to  D,  the  mid- 
point of  the  base. 

Let  the  pupil  supply  the  remainder  of  the  proof. 

Proposition  XXI.  Theorem  (Conv.  of  Prop.  XX) 

800.  If  two  angles  of  a spherical  triangle  are  equal,  the 
sides  opposite  these  angles  are  equal,  and  the  triangle  is 
isosceles. 


V_ 

B'  'C 

Given  the  spherical  A A BC  in  which  ZB—  ZC. 

To  prove  AB  = AC. 

Proof.  Construct  A A’B’C  the  polar  A of  ABC. 

Then  A'C'  = A'B' . Art.  781. 

ZC  = Z B' . Art.  799. 

AB  = AC.  Art.  781. 

Q.  E.  D. 


SPHERICAL  TRIANGLES 


451 


Proposition  XXII.  Theorem 

801.  In  any  spherical  triangle , if  two  angles  are  un- 
equal, the  sides  opposite  these  angles  are  unequal,  and  the 
greater  side  is  opposite  the  greater  angle,  and  Conversely. 


Given  tlie  spherical  A ABC  in  which  ABAC  is  greater 
than  Z G. 

To  prove  BG  > BA. 

Proof.  Draw  the  arc  AB  making  ZDAC  equal  to  Z G. 


Then  DA  = DC.  Art.  800. 

To  each  of  these  equals  add  the  arc  BD. 

:.  BD  + DA  = BD  + DG,  or  BG.  (Why  ») 

But  in  A BDA,  BD  + DA  > BA.  (Why  ?) 

/.  BG  > BA.  Ax.  8. 


Let  the  pupil  prove  the  converse  by  the  indirect  method 
(see  Art.  106) . 

Q.  £.  D. 

Ex.  1.  Bisect  a given  spherical  angle. 

Ex.  2.  Bisect  a given  arc  of  a great  circle  on  a sphere. 

Ex.  3.  At  a given  point  in  an  arc  on  a sphere,  construct  an  angle 
equal  to  a given  spherical  angle  on  the  same  sphere. 

Ex.  4.  Find  the  locus  of  the  centers  of  the  circles  of  a sphere 
formed  by  planes  perpendicular  to  a given  diameter  of  the  given 
sphere. 


452 


BOOK  IX.  SOLID  GEOMETRY 


SPHERICAL  AREAS 

802.  Units  of  spherical  surface.  A spherical  surface 
may  be  measured  in  terms  of,  either 

1.  The  customary  units  of  area , as  a square  inch,  a 
square  foot,  etc.,  or 

2.  Spherical  degrees,  or  spherids. 

803.  A spherical  degree,  or  spherid,  is  one -ninetieth 
part  of  one  of  the  eight  trirectangular  triangles  into  which 
the  surface  of  a sphere  may  be  divided  (Art.  787),  or  t^o 
part  of  the  surface  of  the  entire  sphere. 

A solid  degree  is  one-ninetieth  part  of  a trirectangular  angle  (see 
Art.  774). 

804.  A lune  is  a portion  of  the  surface  of  a sphere 
bounded  by  two  semicircumferences 
of  great  circles,  as  PBP'C  of  Fig.  1. 

The  angle  of  a lune  is  the  angle 
formed  by  the  semicircumferences 
which  bound  it,  as  the  angle  BPC. 

805.  A zone  is  the  portion  of  the 
surface  of  the  sphere  bounded  by 
two  parallel  planes. 

A zone  may  also  be  defined  as  the  sur- 
face generated  by  an  arc  of  a revolving 
semicircumferenee.  Thus,  if  QFQ1  (Fig.  2) 
generates  a sphere  by  rotating  about  QQ' , 
its  diameter,  any  arc  of  QFQ',  as  EF, 
generates  a zone. 

806.  A zone  of  one  base  is  a zone 
one  of  whose  bounding  planes  is 
tangent  to  the  sphere,  as  the  zone 
generated  by  the  arc  QE  of  Fig.  2. 

807.  The  altitude  of  a zone  is  the  perpendicular  dis- 
tance between  the  bounding  planes  of  the  zone. 

The  bases  of  a zone  are  the  circumferences  of  the  circles 

of  the  sphere  formed  by  the  bounding  planes  of  the  zone. 


P 


Q 


SPHERICAL  AREAS 


453 


Proposition  XXIII.  Theorem 

808.  The  area  generated  by  a straight  line  revolving 
about  an  axis  in  its  plane  is  equal  to  the  projection  of  the 
line  upon  the  axis,  multiplied  by  the  circumference  of  a circle 
whose  radius  is  the  perpendicular  erected  at  the  midpoint  of 
the  line  and  terminated  by  the  axis. 


Pig.  1 Fig.  2 Fig.  3 

Given  AB  and  XY  in  the  same  plane,  CD  the  projection 
of  AB  on  XY,  PQ  the  J_  bisector  of  AB-,  and  a surface 
generated  by  the  revolution  of  AB  about  XY,  denoted  as 
"area  AB." 

To  prove  area  AB=  CD  X 2 nPQ. 

Proof.  1.  In  general,  the  surface  generated  by  AB  is 
the  surface  of  a frustum  of  a cone  (Fig.  1). 

area  4B=4B  X 2 nPB.  Art.  728. 

Draw  AF  _L  BD,  then 

A ABF  and  PQR  are  similar.  Art.  328. 

.-.  AB  : AF=PQ  : PR.  (Why?) 

.-.  ABXPR=AFX  PQ,  or  CD  X PQ.  (Why  ?) 

Substituting,  area  AB  — CD  X 2 nPQ.  Ax.  8. 

2.  If  AB\\XY  (Fig.  2),  the  surface  generated  by  AB 
is  the  lateral  surface  of  a cylinder. 

.*.  area  AB  — CD  X 2 nPQ.  Art.  697. 

3.  If  the  point  A lies  in  the  axis  XY  (Fig.  3),  let  the 
pupil  show  that  the  same  result  is  obtained.  q.  e.  d. 


454 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  XXIV.  Theorem 


809.  The  area  of  the  surface  of  a sphere  is  equal  t-o  the 
product  of  the  diameter  of  the  sphere  by  the  circumference 
of  a great  circle. 


Given  a sphere  generated  by  the  revolution  of  the  semi- 
circle AGE  about  the  diameter  AE,  with  the  surface  of  the 
sphere  denoted  by  S,  and  its  radius  by  R. 

To  prove  S=AE  X 2 7tR. 


Proof.  Inscribe  in  the  given  semicircle  the  half  of  a 
regular  polygon  of  an  even  number  of  sides,  as  ABODE. 

Draw  the  apothem  to  each  side  of  the  semipolygon,  and 
denote  it  by  a. 

From  the  vertices  B,  C,  D draw  -b  to  AE. 

Then  area  AB  — AF  X 2 7ia.  1 

area  BC=FO  X 2 7ta.  I 
area  CD=OKX  2 na.  | Art'  808‘ 

area  DE—KE  X 2 na. 


Adding,  area  ABCDE=AE  X 2 na. 


If,  now,  the  number  of  sides  of  the  polygon  be  indefi- 
nitely increased, 

area  ABODE  approaches  $ as  a limit.  Art.  441. 
And  a approaches  A as  a limit.  (Why?) 

.'.  AE  X 2 na  approaches  AE  X 2 nR  as  a limit.  ("Why  ?) 
But  area  ABODE =AE  X 2 na  always. 

S=AE  X 2 nR.  (Why?) 

Q.  £•  ». 


SPHERICAL  AREAS 


455 


810.  Formulas  for  area  of  surface  of  a sphere. 

Substituting  for  AE  its  equal  2 4?,  *S=4  tiR2. 

Also  denoting  the  diameter  of  the  sphere  by  D,  i?  = J D. 

.'.  , or  S=nlF. 

811.  Cor.  1.  The  surface  of  a sphere  is  equivalent  to  four 
times  the  area  of  a great  circle  of  the  sphere. 

812.  Cor.  2.  The  areas  of  the  surfaces  of  two  spheres 
are  to  each  other  as  the  squares  of  their  radii,  or  of  their 
diameters. 

For,  if  S and  S'  denote  the  surfaces,  4?  and  R'  the  radii, 
and  D and  T)’  the  diameters  of  two  spheres, 

S __4:7lR2  R2  S ^P2  B2 

S'  4 71 R'2  R'2'  aiS°S'-7iE'2  E'2' 

813.  Property  of  the  sphere.  The  following  property  of 
the  sphere  is  used  in  the  proof  of  Art.  809:  If,  in  the  generat- 
ing arc  of  any  zone,  a broken  line  be  inscribed,  ivhose  vertices 
divide  the  arc  into  equal  parts,  then,  as  the  number  of  these 
parts  is  increased  indefinitely , the  area  generated  by  the  broken 
line  approaches  the  area  of  the  zone  as  a limit.  Hence 

Cor.  3.  The  area  of  a zone  is  equal  to  the  circumference 
of  a great  circle  multiplied  by  the  altitude  of  the  zone. 

Thus  the  area  generated  by  the  arc  BC  = FO  X 2 71 R. 

814.  Cor:  4.  On  the  same  sphere,  or  on  equal  spheres,  the 
areas  of  two  zones  are  to  each  other  as  the  altitudes  of  the 
zones. 

Ex.  1.  Find  the  area  of  a sphere  whose  diameter  is  10  in. 

Ex.  2.  Find  the  area  of  a zone  of  altitude  3 in.,  on  a sphere  whose 
radius  is  10  in. 


456 


BOOK  IX.  SOLID  GEOMETRY 


Proposition  XXV.  Theorem 

815.  The  area  of  a lime  is  to  the  area  of  the  sur- 
face of  the  sphere  as  the  angle  of  the  lune  is  to  four  right 
angles. 

A 


Given  a sphere  having  its  area  denoted  by  8,  and  on  the 
sphere  the  lune  ABGD  of  A A with  its  area  denoted  by  L. 

To  prove  L : S=X°  : 360°. 

Proof.  Draw  FBH,  the  great  O whose  pole  is  A,  inter- 
secting the  bounding  arcs  of  the  lune  in  B and  D. 

Case  I.  When  the  arc  BT>  and  the  circumference  FBH 
are  commensurable . 

Find  a common  measure  of  BD  and  FBH,  and  let  it  be 
contained  in  the  arc  BD  m times,  and  in  the  circumference 
FBH  n times. 

Then  arc  BD  : circumference  FBH=m  : n. 

Through  the  diameter  AC,  and  the  points  of  division  of 
the  circumference  FBH  pass  planes  of  great  ®. 

The  arcs  of  these  great  © will  divide  the  surface  of  the 
sphere  in  n small  equal  lunes,  m of  them  being  contained 
in  the  lune  ABCD. 

L : S=m  : n. 

L : S= are  BD  : circumference  FBH.  (Why?) 

Or  L : 8=A°  : 360°.  Art.  257. 

Case  II.  When  the  arc  BD  and  the  circumference  FBH 
are  incommensurable . 

Let  the  pupil  supply  the  proof. 


Q.  E.  D. 


SPHEEICAL  AREAS 


457 


816.  Formula  for  the  area  of  a lune  in  spherical  degrees, 
or  spherids.  The  surface  of  a sphere  contains  720  spherids 
(Art.  803).  Hence,  by  Art.  815, 


L L spherids  _ A 
8 01  720  spherids  360’ 


^ = A,  or  L = 2 A spherids; 


that  is,  the  area  of  a lune  in  spherical  degrees  is  equal  to 
twice  the  number  of  angular  degrees  in  the  angle  of  the  lune. 


817.  Formula  for  area  of  a lune  in  square  units  of  area. 

0 . x L _ A T_KR2A 

S — 4-  TtR*  (Art.  810)  ..4  -360,  or  L 90 

818.  Cor.  1.  On  the  same  sphere,  or  on  equal  spheres , 
two  tunes  are  to  each  other  as  their  angles. 


819.  Cor.  2.  Two  limes  with  equal  angles,  but,  on  un- 
equal spheres,  are  to  each  other  as  the  squares  of  the  radii  of 
their  spheres. 


For  L : L'  = 


TtRfA 

90 


TtR^A 
''  90  ’ 


or  L : L'  — R2  : R'2. 


Ex.  1.  Find  the  area  in  spherical  degrees  of  a lune  of  27°. 

Ex.  2.  Find  the  number  of  square  inches  in  the  area  of  a lune  of 
27°,  on  a sphere  whose  radius  is  10  in. 

A solid  symmetrical  with  respect  to  a plane  is  a solid  in  which 
a line  drawn  from  any  point  in  its  surface  X the  given  plane  and 
produced  its  own  length  ends  in  a point  on  the  surface ; hence 

Ex.  3.  How  many  planes  of  symmetry  has  a circular  cylinder  ? A 
cylinder  of  revolution  ? 

Ex.  4.  Has  either  of  these  solids  a center  of  symmetry  ? 

Ex.  5.  Answer  the  same  questions  for  a circular  cone. 

Ex.  6.  For  a cone  of  revolution.  For  a sphere. 

Ex.  7.  For  a regular  square  pyramid.  For  a regular  pentagonal 
pyramid. 


458 


BOOK  IX.  SOLID  GEOMETEY 


Proposition  XXVI.  Theorem 

820.  If  two  great  circles  intersect  on  a hemisphere,  the 
sum  of  two  vertical  triangles  thus  formed  is  equivalent  to  a 
lune  ivhose  angle  is  that  angle  in  the  triangles  which  is 
formed  by  the  intersection  of  the  two  great  circles. 


Given  the  hemisphere  ABBF,  and  on  it  the  great  circles 
AFB  and  I)FC,  intersecting  at  F. 

To  prove  A AFC  + A BFD  = lune  whose  Z is  BFD. 

Proof.  Complete  the  sphere  and  produce  the  given  arcs 
of  the  great  circles  to  intersect  at  F'  on  the  other  hemisphere. 

Then,  in  the  A AFC  and  BF'B, 

arc  ZL.F=arc  BF' , 

( each  being  the  supplement  of  the  arc  BF). 

In  like  manner  arc  CF=  arc  BF' . 

And  arc  A G—  arc  BB. 

A AFCo  A BF'B.  Art.  795. 

Add  the  A BFB  to  each  of  these  equals; 

.-.  A AFC  + A BFB^A  BF'B  + A BFB.  Ax.  3. 

Or  /.  A AFC  A-  A BFB^lxme  FBF'B.  Ax.  6. 

Q.  E.  D. 


459 


SPHERICAL  AREAS 

Proposition  XXVII.  Theorem 

821.  The  number  of  spherical  degrees,  or  spherids,  in 
the  area  of  a spherical  triangle  is  equal  to  the  number  of 
angular  degrees  in  the  spherical  excess  of  the  triangle. 


Given  the  spherical  A ABC  whose  A are  denoted  by  A, 
B,  C , and  whose  spherical  excess  is  denoted  by  E. 

To  prove  area  of  A ABC  — E spherids. 

Proof.  Produce  the  sides  AC  and  BC  to  meet  AB  pro- 
duced in  the  points  D and  F,  respectively. 

A ABC  A A CDB  = luue  ABDC  — 2 A spherids.  ) 

r Art.  816. 

A ABC  A A ACE  =lune  BCFA  — 2 B spherids.  j 

A ABC  A A CFB  = lune  of  aBCA  = 2 C spherids.  Art.  820- 

Adding,  and  observing  that  A ABCAA  CDB  A AACF 
-f-  A CFD  = hemisphere  ABDFC, 

2 A ABC  A hemisphere  = 2 (A  + B + C)  spherids.  Or, 
2 A ABC  + 360  spherids  = 2 {A  + B + C)  spherids.  Ax.  8. 

A ABC  + 180  spherids=  (A  AB  + C)  spherids.  Ax.  5. 
A ABC=  (A  A B A C — 180)  spherids.  Ax.  3. 
Or  area  A ABC—E  spherids. 


Art.  788. 

Q.  E.  D. 


460 


BOOK  IX.  SOLID  GEOMETRY 


822,  Formula  for  area  of  a spherical  triangle  in  square 
units  of  area. 

Comparing  the  area  of  a spherical  A with  the  area  of 
the  entire  sphere , 

area  A : 4 7t R2  = E spherids  : 720  spherids. 

A 4 tie2  X E A 7iB2E 

.-.  area  A =s — , or  area  A=— ■ 

823.  The  spherical  excess  of  a spherical  polygon  is  the 

sum  of  the  angles  of  the  polygon  diminished  by  (« — 2)  180°; 
that  is,  it  is  the  sum  of  the  spherical  excesses  of  the  tri- 
angles into  which  the  polygon  may  be  divided. 


Proposition  XXVIII.  Theorem 

824.  The  area  of  a spherical  polygon , in  spherical  de- 
grees or  spherids , is  equal  to  the  spherical  excess  of  the 
polygon.  a 


Given  a spherical  polygon  ABCJDF  of  n sides,  with  its 
spherical  excess  denoted  by  E. 

To  prove  area  of  ABODE— E spherical  degrees. 

Proof.  Draw  diagonals  from  A,  any  vertex  of  the  poly- 
gon, and  thus  divide  the  polygon  into  n — 2 spherical  A. 

The  area  of  each  A = (sum  of  its  A — 180)  spherids. 

Art.  821. 

.*.  sum  of  the  areas  of  the  A = [sum  of  A of  the  A — 
(» — 2)  180]  spherids.  Ax.  2. 

.'.  area  of  polygon  = E spherids,  Art.  823. 

( for  the  sum  of  A of  the  A — {n — 2)  180°  =E°), 


SPHERICAL  VOLUMES 


4G1 


SPHERICAL  VOLUMES 

825.  A spherical  pyramid  is  a por- 
tion of  a sphere  bounded  by  a spheri- 
cal polygon  and  the  planes  of  the 
great  circles  forming  the  sides  of  the 
polygon.  The  base  of  a spherical 
pyramid  is  the  spherical  polygon 
bounding  it,  and  the  vertex  of  the 
spherical  pyramid  is  the  center  of  the  sphere. 

Thus,  in  the  spherical  pyramid  O-ABCD,  the  base  is 
ABCD  and  the  vertex  is  0. 

826.  A spherical  wedge  (or  ungula)  is  the  portion  of  a 
sphere  bounded  by  a lune  and  the  planes  of  the  sides  of 
the  lune. 


827.  A spherical  sector  is  the  portion  of  a sphere  gene- 
rated by  a sector  of  that  semicircle  whose  rotation  generates 
the  given  sphere. 


828.  The  base  of  a spherical  sector  is  the  zone  gene- 
rated by  the  revolution  of  the  arc  of  the  plane  sector  which 
generates  the  spherical  sector. 

Let  the  pupil  draw  a spherical  sector  in  which  the  base 
is  a zone  of  one  base. 


462 


BOOK  IX.  SOLID  GEOMETRY 


829.  A spherical  segment  is  a portion  of  a sphere 
included  between  two  parallel  planes. 

The  bases  of  a spherical  segment  are  the  sections  of  the 
sphere  made  by  the  parallel  planes  which  bound  the  given 
segment;  the  altitude  is  the  perpendicular  distance  between 
the  bases. 

830.  A spherical  segment  of  one  base  is  a spherical  seg- 
ment one  of  whose  bounding  planes  is  tangent  to  the 
sphere. 


Proposition  XXIS.  Theorem 

831.  The  volume  of  a sphere  is  equal  to  one-third  the 
product  of  the  area  of  its  surface  by  its  radius. 


Given  a sphere  having  its  volume  denoted  by  V,  sur« 
face  by  S,  and  radius  by  B. 

To  prove  Y=$SXR. 

Proof.  Let  any  polyhedron  be  circumscribed  about  the 
sphere. 

Pass  a plane  through  each  edge  of  the  polyhedron  and 
the  center  of  the  sphere. 

These  planes  will  divide  the  polyhedron  into  as  many 
pyramids  as  the  polyhedron  has  faces,  each  pyramid  hav- 
ing a face  of  the  polyhedrou  for  its  base,  the  center  of  the 


SPHERICAL  VOLUMES 


463 


sphere  for  its  vertex,  and  the  radius  of  the  sphere  for  its 
altitude. 

.'.  volume  of  each  pyramid  = £ base  X R.  (Why  ?) 

.*.  volume  of  polyhedron  = £ (surface  of  polyhedron)  XR. 

If  the  number  of  faces  of  the  polyhedron  be  increased 
indefinitely,  the  volume  of  the  polyhedron  approaches  the 
volume  of  the  sphere  as  a limit,  and  the  surface  of  the 
polyhedron  approaches  the  surface  of  the  sphere  as  a 
limit. 

Hence  the  volume  of  the  polyhedron  and  J (surface  of 
the  polyhedron)  X R,  are  two  variables  always  equal. 

Hence  their  limits  are  equal, 

Or  V=$SXR.  (Why?) 

Q.  E.  D. 


832.  Formulas  for  volume  of  a sphere.  Substituting 
$=4  7tR2,  or  S = 7l-D"  in  the  result  of  Art.  831, 


V- 


4 n r? 
3 ; 


7 tD3 

also  V=—~r~- 
b 


833.  Cor.  1.  The  volumes  of  two  spheres  are  to  each 
other  as  the  cubes  of  their  radii , or  as  the  cubes  of  their 
diameters. 

V _ J TtR3  _R3  Y j 7iJ>3  H3 

*°r  V'  i 7iR'3  R /3;  aiS0  V’  in  D'3  I)'3’ 


834.  Cor.  2.  The  volume  of  a spherical  pyramid  is 
equal  to  one- third  the  product  of  its  base  by  the  radius  of 
the  sphere. 

835.  Cor.  3.  The  volume  of  spherical  sector  is  equal  to 
one-third  the  product  of  its  base  ( the  bounding  zone)  by  the 
radius  of  the  sphere. 


464 


BOOK  IX.  SOLID  GEOMETRY 


836.  Formula  for  the  volume  of  a spherical  sector.  De- 
noting the  altitude  of  the  sector  by  H and  the  volume  by  V, 

V—  £ (area  of  zone)  X E, 

= £ (2  nER)  R.  Art.  813. 

.*.  V=‘%  nE2E. 

Proposition  XXX.  Theorem 

837.  The  volume  of  a spherical  segment  is  equal  to  one- 
half  the  product  of  its  altitude  by  the  sum  of  the  areas  of 
its  bases,  plus  the  volume  of  a sphere  tvhose  diameter  is  the 
altitude  of  the  segment. 


A 


Given  the  semicircle  ABCA'  which  generates  a sphere 
by  its  rotation  about  the  diameter  AA' ; BT>  and  OP1  semi- 
chords _L  AA' , and  denoted  by  r and  r'\  and  BF  denoted 
by  H. 

To  prove  volume  of  spherical  segment  generated  by 
BCFD,  or  F=£(7ir2  + 7tr'2)  HA  I nH3. 

Proof.  Draw  the  radii  OB  and  00. 

Denote  OF  by  h,  and  OB  by  h. 

Then  F=vol.  OBC  A x ol.  OCF — vol.  OBB. 

:.  F=|  7lR2H  + £ nr'2h — £ Itr-h.  Arts.  836,  723. 

But  H=  h-k,  Jr = Rt-r'2,  and  k2 = B~r2,  (Why  ?) 


SPHERICAL  VOLUMES 


465 


V=i  7 1 [2  B2  (7 HfeH  (S2-/^)  h-iB2-^)  k\ . Ax.  8. 
= * 7t  [2  B2  (h-k)  + B2  (h-k)-(h3-k3)']. 

= i 7t  E [3  £2— (7r  + M + &2)]  . 

But  7*2 — 2 7ift  Jrk2=E2.  Ax.  4. 

Subtract  each  member  from  3 7t2  + 3 k2  and  divide  by  2. 

Axs.  3,  5. 

Then  h2  + lik  + k2= I (7 i2  + k2)-^f- 

= 3 B2 — 1 (r2  + r'2)~- 

V=i  nE  ft  (r2  + r/2)  + ~ ~|  • Ax.  8. 

A V=i  Ur2+7ir'2)  E+inE3. 

Q.  E.  D. 

838.  Formula  for  volume  of  a spherical  segment  of  one 
base.  In  a spherical  segment  of  one  base  v'  = o,  and  r2  = 
(2  B-E)  E (Art.  343). 

Substituting  for  r and  r'  these  values  in  the  result  of 
Art.  837, 

F=*ff2(HD- 

839.  Advantage  of  measurement  formulas.  The  student 

should  observe  carefully  that,  by  the  results  obtained  in 
Book  15,  the  measurement  of  the  areas  of  certain  curved 
surfaces  is  reduced  to  the  far  simpler  work  of  the  measure- 
ment of  the  lengths  of  one  or  more  straight  lines;  in  like 
manner  the  measurement  of  certain  volumes  bounded  by  a 
curved  surface  is  reduced  to  the  simpler  work  of  linear 
measurements.  A similar  remark  applies  to  the  results  of 
Book  VIII.  

Ex.  1.  Find  the  volume  of  a sphere  whose  radius  is  7 in. 

Ex.  2.  Find  the  volume  of  a sphere  whose  diameter  is  7 in. 

Ex.  3.  In  a sphere  whose  radius  is  8 in.,  find  the  volume  of  a 
spherical  segment  of  one  base  whose  altitude  is  3. 

DD 


466 


BOOK  IX.  SOLID  GEOMETRY 


EXERCISES.  CROUP  72 

THEOREMS  CONCERNING  THE  SPHERE 

Ex.  1.  Of  circles  of  a sphere  whose  planes  pass  through  a given 
point  within  a sphere,  the  smallest  is  that  circle  whose  plane  is 
perpendicular  to  the  diameter  through  the  given  point. 

Ex.  2.  If  a point  on  the  surface  of  a given  sphere  is  equidistant 
from  three  points  on  a given  small  circle  of  the  sphere,  it  is  the  pole 
of  the  small  circle. 

Ex.  3.  If  two  sides  of  a spherical  triangle  are  quadrants,  the  third 
side  measures  the  angle  opposite  that  side  in  the  triangle. 

Ex.  4.  If  a spherical  triangle  has  one  right  angle,  the  sum  of  its 
other  two  angles  is  greater  than  one  right  angle. 

Ex.  5.  If  a spherical  triangle  is  isosceles,  its  polar  triangle  is 
isosceles. 

Ex.  6.  The  polar  triangle  of  a birectangular  triangle  is  birectan- 
gular. 

Ex.  7.  The  polar  triangle  of  a trirectangular  triangle  is  identical 
with  the  original  triangle. 

Ex.  8.  Prove  that  the  sum  of  the  angles  of  a spherical  quadri- 
lateral is  greater  than  4 right  angles,  and  less  than  8 right  angles. 
What,  also,  are  the  limits  of  the  sum  of  the  angles  of  a spherical 
hexagon  ? Of  the  sum  of  the  angles  of  a spherical  «-gon  ? 

Ex.  9.  On  the  same  sphere,  or  on  equal  spheres,  two  birectan- 
gular triangles  are  equal  if  their  oblique  angles  are  equal. 

Ex.  10.  Two  zones  on  the  same  sphere,  or  on  equal  spheres,  are 
to  each  other  as  their  altitudes. 

Ex.  11.  If  one  of  the  legs  of  a right  spherical  triangle  is  greater 
than  a quadrant,  another  side  is  also  greater  than  a quadrant. 

[Sug.  Of  the  leg  which  is  greater  than  a quadrant,  take  the  end 
remote  from  the  right  angle  as  a pole,  and  describe  an  arc.] 

Ex.  12.  It  ABC  and  A'B'C  are  polar  triangles,  the  radius  OA  is 
perpendicular  to  the  plane  OB'C'. 


EXERCISES  ON  THE  SPHERE 


467 


Ex.  13.  On  the  same  sphere,  or  on  equal  spheres,  spherical  tri- 
angles whose  polar  triangles  have  equal  perimeters  are  equivalent. 

Ex.  14.  Given  OAO',  OBO',  and  AB  arcs  of  great 
circles,  intersecting  so  that  Z.OAB=  Z O’BA  ; prove  that 
/AOAB=/AO’AB. 

[Sue.  Show  that  Z OBA=  Z 0'AB.~\ 

Ex.  1 5.  Find  the  ratio  of  the  volume  of  a sphere  to 
the  volume  of  a circumscribed  cube. 

Ex.  16.  Find  the  ratio  of  the  surface  of  a spnere  to 
the  lateral  surface  of  a circumscribed  cylinder  of  revolution;  also 
find  the  ratio  of  their  volumes. 

Ex.  17.  If  the  edge  of  a regular  tetrahedron  is  denoted  by  a,  find 
the  ratio  of  the  volumes  of  the  inscribed  and  circumscribed  spheres. 

Ex.  18.  Find  the  ratio  of  the  two  segments  into  which  a hemi- 
sphere is  divided  by  a plane  parallel  to  the  base  of  the  hemisphere  and 
at  the  distance  from  the  base. 


EXERCISE8.  GROUP  73 

SPHERICAL  LOCI 

Ex.  1.  Find  the  locus  of  a point  at  a given  distance  a from  the 
surface  of  a given  sphere. 

Ex.  2.  Find  the  locus  of  a point  on  the  surface  of  a sphere  that  is 
equidistant  from  two  given  points  on  the  surface. 

Ex.  3.  If,  through  a given  point  outside  a given  sphere,  tangent 
planes  to  the  sphere  are  passed,  find  the  locus  of  the  points  of 
tangency. 

Ex.  4.  If  straight  lines  be  passed  through  a given  fixed  point  in 
space,  and  through  another  given  point  other  straight  lines  be  passed 
perpendicular  to  the  first  set,  find  the  locus  of  the  feet  of  the 
perpendiculars. 


468 


BOOK  IX.  SOLID  GEOMETRY 


EXERCISES.  CROUP  74 

PROBLEMS  CONCERNING  THE  SPHERE 

Ex.  1.  At  a given  point  on  a sphere,  construct  a plane  tangent  to 
the  sphere. 

Ex.  2.  Through  a given  point  on  the  surface  of  a sphere,  draw  an 
arc  of  a great  circle  perpendicular  to  a given  arc. 

Ex.  3.  Inscribe  a circle  in  a given  spherical  triangle. 

Ex.  4.  Construct  a spherical  triangle,  given  its  polar  triangle. 

Given  the  radius,  r,  construct  a spherical  surface  which  shall  pass 
through 

Ex.  5.  Three  given  points. 

Ex.  6.  Two  given  points  and  be  tangent  to  a given  plane. 

Ex.  7.  Two  given  points  and  bQ  tangent  to  a given  sphere. 

Ex.  8.  One  given  point  and  be  tangent  to  two  given  planes. 

Ex.  9.  One  given  point  and  be  tangent  to  two  given  spheres. 

Given  the  radius,  r,  construct  a spherical  surface  which  shall  be 
tangent  to 

Ex.  10.  Three  given  planes. 

Ex.  11.  Two  given  planes  and  one  given  sphere. 

Ex.  12.  Construct  a spherical  surface  which  shall  pass  through 
three  given  points  and  be  tangent  to  a given  plane. 

Ex.  13.  Through  a given  straight  line  pass  a plane  tangent  to  a 
given  sphere. 

[Sug.  Through  the  center  of  the  sphere  pass  a plane  J.  given 
line,  etc.] 

When  is  the  solution  impossible  ? 

Ex.  14.  Through  a given  point  on  a sphere,  construct  an  are  of 
a great  circle  tangent  to  a given  small  circle  of  the  sphere. 

[Sug.  Draw  a straight  line  from  the  center  of  the  sphere  to  the 
given  point,  and  produce  it  to  intersect  the  plane  of  the  small  circle, 
etc.] 


NUMERICAL  EXERCISES  IN  SOLID 
GEOMETRY 


For  methods  of  facilitating  numerical  computations, 
see  Arts.  493-6. 


EXERCISES.  CROUP  75 

LINES  AND  SURFACES  OF  POLYHEDRONS 
find  the  lateral  area  and  total  area  of  a right  prism  whose 

Ex.  1.  Base  is  an  equilateral  triangle  of  edge  4 in.,  and  whose 
altitude  is  15  in. 

Ex.  2.  Base  is  a triangle  of  sides  17,  12,  25,  and  whose  altitude 
is  20. 

Ex.  3.  Base  is  an  isosceles  trapezoid,  the  parallel  bases  being  10 
and  15  and  leg  8,  and  whose  altitude  is  24. 

Ex.  4.  Base  is  a rhombus  whose  diagonals  are  12  and  16,  and 
whose  altitude  is  12. 

Ex.  5.  Base  is  a regular  hexagon  with  side  8 ft.,  and  whose  alti- 
tude is  20  ft. 

Ex.  6.  Find  the  entire  surface  of  a rectangular  parallelopiped 
8 X 12  X 16  in- ; of  one  p X q X r ft. 

Ex.  7.  Of  a cube  whose  edge  is  1 ft.  3 in. 

Ex.  8.  The  lateral  area  of  a regular  hexagonal  prism  is  120  sq.  ft. 
and  an  edge  of  the  base  is  10  ft.  Find  the  altitude. 

Ex.  9.  How  many  square  feet  of  tin  are  necessary  to  line  a box 
20 X 6X4  in.  ? 

Ex.  10.  If  the  surface  of  a cube  is  1 sq.  yd. , find  an  edge  in  inches. 

Ex.  11.  Find  the  diagonal  of  a cube  whose  edge  is  5 in. 

Ex.  12.  If  the  diagonal  of  a cube  is  12  ft.,  find  the  surface. 

(469) 


470 


'solid  geometry 


Ex.  13.  If  the  surface  of  a rectangular  parallelopiped  is  208  sq.  in., 
and  the  edges  are  as  2 : 3 : 4,  find  the  edges. 

In  a regular  square  pyramid 

V Ex.  14.  If  an  edge  of  the  base  is  16  and  slant  height  is  17,  find 
the  altitude. 

Ex.  15.  If  the  altitude  is  15  and  a lateral  edge  is  17,  find  an  edge 
of  the  base. 

Ex.  16.  If  a lateral  edge  is  25  and  an  edge  of  the  base  is  14,  find 
the  altitude. 

In  a regular  triangular  pyramid 

Ex.  17.  If  an  edge  of  the  base  is  8 and  the  altitude  is  10,  find  the 
slant  height. 

Ex.  18.  Find  the  altitude  of  a regular  tetrahedron  whose  edge  is  6. 

Find  the  lateral  surface  and  the  total  surface  of 

Ex.  19.  A regular  square  pyramid  an  edge  of  whose  base  is  16, 
and  whose  altitude  is  15. 

Ex.  20.  A regular  triangular  pyramid  an  edge  of  whose  base  is  10, 
and  whose  altitude  is  12. 

Ex.  21.  A regular  hexagonal  pyramid  an  edge  of  whose  base  is  4, 
and  whose  altitude  is  21. 

Ex.  22.  A regular  square  pyramid  whose  slant  height  is  24,  and 
•whose  lateral  edge  is  25. 

Ex.  23.  A regular  tetrahedron  whose  edge  is  4. 

Ex.  24.  A regular  tetrahedron  whose  altitude  is  9. 

Ex.  25.  A regular  hexagonal  pyramid  each  edge  of  whose  base  is 
a,  and  whose  altitude  is  b. 

Ex.  26.  In  the  frustum  of  a regular  square  pyramid  the  edges  of 
the  bases  are  6 and  18,  and  the  altitude  is  8.  Find  the  slant  height. 
/Hence  find  the  lateral  area. 

Ex.  27.  In  the  frustum  of  a regular  triangular  pyramid  the  edges 
' of  the  bases  are  4 and  6,  and  the  altitudi  is  5,  Find  the  slant  height. 
Hence  find  the  lateral  area. 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  471 


Ex.  28.  In  the  frustum  of  a regular  tetrahedron,  if  the  edge  of  the 
lower  base  is  &i,  the  edge  of  the  upper  base  is  bo,  and  the  altitude  is  a, 
show  that  Z=i}/ } ( &i  — bi)2  + 4 a1. 

Ex.  29.  In  the  frustum  of  a regular  square  pyramid  the  edges  of 
the  bases  are  20  and  60,  and  a lateral  edge  is  101.  Find  the  lateral 
surface. 


EXERCISES.  CROUP  76 

LINES  AND  SURFACES  OF  CONES  AND  CYLINDERS 

Ex.  1.  How  many  square  feet  of  lateral  surface  has  a tunnel  100 
yds.  long  and  7 ft.  in  diameter. 

Ex.  2.  The  lateral  area  of  a cylinder  of  revolution  is  1 sq.  yd., 
and  the  altitude  is  1 ft.  Find  the  radius  of  the  base. 

Ex.  3.  The  entire  surface  of  a cylinder  of  revolution  is  900  sq.  ft. 
and  the  radius  of  the  base  is  10  ft.  Find  the  altitude. 

In  a cylinder  of  revolution 

Ex.  4.  Find  B in  terms  of  S and  R. 

Ex.  5.  Find  R in  terms  of  B and  T. 

Ex.  6.  Find  Tin  terms  of  S and  R. 

Ex.  7.  How  many  sq.  yds  of  canvas  are  required  to  make  a coni- 
cal tent  20  ft.  in  diameter  and  12  ft.  high  f 

Ex.  8.  A man  has  400  sq.  yds.  of  canvas  and  wants  to  make  a 
conical  tent  20  yds.  in  diameter.  What  will  be  its  altitude  ? 

Ex.  9.  The  altitude  of  a cone  of  revolution  is  10  ft.  and  the  lat- 
eral area  is  11  times  the  area  of  the  base.  Find  the  radius  of  the  base. 

In  a cone  of  revolution 

Ex.  10.  Find  T in  terms  of  S and  L. 

Ex.  11.  Find  B in  terms  of  T and  L. 

Ex.  12.  How  many  square  feet  of  tin  are  necessary  to  make  a 
funnel  the  diameters  of  whose  ends  are  2 in.  and  8 in.,  and  whose 
altitude  is  7 in.  ? 

Ex.  13.  If  the  slant  height  of  a frustum  of  a cone  of  revolution 
makes  an  angle  of  45°  with  the  base,  show  that  the  lateral  area  of  the 
frustum  is  (rj2 — r22)  wj/2. 


472 


SOLID  GEOMETRY 


EXERCI8ES.  CROUP  77 

SPHERICAL  LINES  AND  SURFACES 

Ex.  1.  Find  in  square  feet  the  area  of  the  surface  of  a sphere 
whose  radius  is  1 ft.  2 in. 

Ex.  2.  How  many  square  inches  of  leather  will  it  take  to  cover  a 
baseball  whose  diameter  is  in.? 

Ex.  3.  How  many  sq.  ft.  of  tin  are  required  to  cover  a dome  in 
the  shape  of  a hemisphere  6 yds.  in  diameter  ? 

Ex.  4.  What  is  the  radius  of  a sphere  whose  surface  is  616  sq.  in.  ? 

Ex.  5.  Find  the  diameter  of  a globe  whose  surface  is  1 sq.  yd. 

Ex.  6.  If  the  circumference  of  a great  circle  on  a sphere  is  1 ft., 
find  the  area  of  the  surface  of  the  sphere. 

Ex.  7.  If  a hemispherical  dome  is  to  contain  100  sq.  yds.  of  sur- 
face, what  must  its  diameter  be  1 

Ex.  8.  Find  the  radius  of  a sphere  in  which  the  area  of  the  sur- 
face equals  the  number  of  linear  units  in  the  circumference  of  a great 
circle. 

Find  the  area  of  a lune  in  which 

Ex.  9.  The  angle  of  the  lune  is  36°,  and  the  radius  of  the  sphere 
is  14  in. 

Ex.  10.  The  angle  of  the  lune  is  18°  20',  and  the  diameter  of  the 
sphere  is  20  in. 

Ex.  11.  The  angle  of  the  lune  is  24°,  and  the  surface  of  the 
sphere  is  4 sq  ft. 

Find  the  area  of  a spherical  triangle  in  which 

Ex.  12.  The  angles  are  80°,  90°,  120°,  and  the  diameter  of  the 
sphere  is  14  ft. 

Ex.  13.  The  angles  are  74°  24',  83°  16',  92°  20',  and  the  radius  of 
the  sphere  is  10. 

Ex.  14.  The  angles  are  85°,  95°,  135°,  and  the  surface  of  the 
sphere  is  10  sq.  ft. 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  473 


Ex.  15.  If  the  sides  of  a spherical  triangle  are  100°,  110°,  120° 
and  the  radius  of  the  sphere  is  16,  find  the  area  of  the  polar  triangle. 

Ex.  16.  If  the  angles  of  a spherical  triangle  are  90°,  100°,  120® 
and  its  area  is  3900,  find  the  radius  of  the  sphere. 

Ex.  17.  If  the  area  of  an  equilateral  spherical  triangle  is  one- 
third  the  surface  of  the  sphere,  find  an  angle  of  the  triangle. 

Ex.  18.  In  a trihedral  angle  the  plane  angles  of  the  dihedral 
angles  are  80°,  90°,  100°;  find  the  number  of  solid  degrees  in  the 
trihedral  angle. 

Ex.  19.  Find  the  area  of  a spherical  hexagon  each  of  whose 
angles  is  150°,  on  a sphere  whose  radius  is  20  in. 

Ex.  20.  If  each  dihedral  angle  of  a given  pentahedral  angle  is 
120°,  how  many  solid  degrees  does  the  pentahedral  angle  contain  ? 

Ex.  21.  In  a sphere  whose  radius  is  14  in.,  find  the  area  of  a zone 
3 in.  high. 

Ex.  22.  What  is  the  area  of  the  north  temperate  zone,  if  the 
earth  is  taken  to  be  a sphere  with  a radius  of  4,000  miles,  and  the 
distance  between  the  plane  of  the  arctic  circle  and  that  of  the  tropic 
of  Cancer  is  1,800  miles  ? 

Ex.  23.  If  Cairo,  Egypt,  is  in  latitude  30°,  show  that  its  parallel 
of  latitude'  bisects  the  surface  of  the  northern  hemisphere. 

Ex.  24.  How  high  must  a person  be  above  the  earth’s  surface  to 
see  one -third  of  the  surface  ? 

Ex.  25.  How  much  of  the  earth’s  surface  will  a man  see  who  is 
2,000  miles  above  the  surface,  if  the  diameter  is  taken  as  8,000  miles? 

Ex.  26.  If  the  area  of  a zone  equals  the  area  of  a great  circle, 
find  the  altitude  of  the  zone  in  terms  of  the  radius  of  the  sphere. 

Ex.  27.  If  sounds  from  the  Krakatoa  explosion  were  heard  at  a 
distance  of  3,000  miles  (taken  as  a chord)  on  the  surface  of  the  earth, 
over  what  fraction  of  the  earth’s  surface  were  they  heard  ? 

Ex.  28.  The  radii  of  two  spheres  are  5 and  12  in.  and  their  cen- 
ters are  13  in.  apart.  Find  the  area  of  the  circle  of  intersection  and 
also  of  that  part  of  the  surface  of  each  sphere  not  included  by  the 
other  sphere. 


474 


SOLID  GEOMETRY 


EXERCISES.  CROUP  78 

VOLUMES  OF  POLYHEDRONS 
r Find  the  volume  of  a prism 

Ex.  1.  Whose  base  is  au  equilateral  triangle  with  side  5 in.,  and 
whose  altitude  is  16  in. 

J Ex.  2.  Whose  base  is  a triangle  with  sides  12,  13,  15,  and  whose 
altitude  is  20. 

1 Ex.  3.  Whose  base  is  an  isosceles  right  triangle  with  a leg  equal 
to  2 yds.,  and  whose  altitude  is  25  ft. 

Ex.  4.  Whose  base  is  a regular  hexagon  with  a side  of  8 ft.,  and 
^ whose  altitude  is  10  yds. 


J Ex.  5.  Whose  base  is  a rhombus  one  of  whose  sides  is  25,  and  one 
^ vA  ^o^  whose  diagonals  is  14,  and  whose  altitude  is  11. 

' * ° 

££*  q Ex.  6.  Whose  base  contains  84  sq.  yds.,  and  whose  lateral  faces 
& ) are  three  rectangles  with  areas  of  100,  170,  210  sq.  yds.,  respectively. 


^ J a're  three 
\ Ex.  7. 


How  many  bushels  of  wheat  are  held  by  a bin  30  x 10  x 6 ft. , 
if  a bushel  is  taken  as  lj  cu.  ft.? 

’J  Ex.  8.  How  many  cart-loads  of  earth  are  in  a cellar  30  x 20x6  ft., 
if  a cart-load  is  a cubic  yard  ? 

'Iex.  9.  If  a cubical  block  of  marble  costs  $2,  what  is  the  cost  of 
a cube  whose  edge  is  a diagonal  of  the  first  block  ? 

J Ex.  10.  Find  the  edge  of  a cube  whose  volume  equals  the  sum  of 
the  volumes  of  two  cubes  whose  edges  are  3 and  5 ft. 

^ Ex.  11.  Find  the  edge  of  a cube  whose  volume  equals  the  area  of 
its  surface. 


Ex.  12.  If  the  top  of  a cistern  is  a rectangle  12x8  ft.,  how  deep 
ust  the  cistern  be  to  hold  10,000  gallons  ? 


Ex.  13.  Find  the  inner  edge  of  a peck  measure  which  is  in  the 
shape  of  a cube. 

Ex.  14.  A peck  measure  is  to  be  a rectangular  parallelopiped  with 
square  base  and  altitude  equal  to  twice  the  edge  of  the  base.  Find 
its  dimensions. 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  475 


Ex.  15.  Find  the  volume  of  a cube  whose  diagonal  is  a. 

Find  the  volume  of  a pyramid 

Ex.  16.  Whose  base  is  an  equilateral  triangle  with  side  8 in., 
and  whose  altitude  is  12  in. 

Ex.  17.  Whose  base  is  a right  triangle  with  hypotenuse  29  and 
one  leg  21,  and  whose  altitude  is  20. 

Ex.  18.  Whose  base  is  a square  with  side  6,  and  each  of  whose 
lateral  edges  is  5. 

Ex.  19.  Whose  base  is  a square  with  side  10,  and  each  of  whose 
lateral  faces  makes  an  angle  of  45°  with  the  base. 

Ex.  20.  If  the  pyramid  of  Memphis  has  an  altitude  of  146  yds. 
and  a square  base  of  side  232  yds.,  how  many  cubic  yards  of  stone 
does  it  contain  ? What  is  this  worth  at  $1  a cu.  yd.  ? 

Ex.  21.  A church  spire  150  ft.  high  is  hexagonal  in  shape  and 
each  side  of  the  base  is  10  ft.  The  spire  has  a hollow  hexagonal 
interior,  each  side  of  whose  base  is  6 ft.,  and  whose  altitude  is  45  ft. 
How  many  cubic  yards  of  stone  does  the  spire  contain  ? 

Ex.  22.  If  a pyramid  contains  4 cu.  yds.  and  its  base  is  a square 
with  one  side  2 ft.,  find  the  altitude. 

Ex.  23.  A heap  of  candy  in  the  shape  of  a frustum  of  a regular 
square  pyramid  has  the  edges  of  its  bases  25  and  9 in.  and  its  altitude 
12  in.  Find  the  number  of  pounds  in  the  heap  if  a pound  is  a rectan- 
gular parallelopiped  4x3  x 2 in.  in  size. 

Ex.  24.  Find  the  volume  of  a frustum  of  a regular  triangular 
pyramid,  the  edges  of  the  bases  being  2 and  8,  and  the  slant  height  12. 

Ex.  25.  The  edges  of  the  bases  of  the  frustum  of  a regular  square 
pyramid  are  24  and  6,  and  each  lateral  edge  is  15 ; find  the  volume. 

Ex.  26.  If  a stick  of  timber  is  in  the  shape  of  a frustum  of  a 
regular  square  pyramid  with  the  edges  of  its  ends  9 and  15  in.,  and 
with  a length  of  14  ft.,  find  the  number  of  feet  of  lumber  in  the  stick. 

What  is  the  difference  between  this  volume  and  that  of  a stick  of 
the  same  length  having  the  shape  of  a prism  with  a base  equal  to  the 
area  of  a midsection  of  the  first  stick  ? 


476 


SOLID  GEOMETRY 


Ex.  27.  Find  the  volume  of  aprismoid  whose  bases  are  rectangles 
5 x 2 ft.  and  7x4  ft.,  and  whose  altitude  is  12  ft. 

Ex.  28.  How  many  cart-loads  of  earth  are  there  in  a railroad  cut 
12  ft.  deep,  whose  base  is  a rectangle  100x8  ft.,  and  whose  top  is  a 
rectangle  30x50  ft.? 

Ex.  29.  Find  the  volume  of  a prismatoid  whose  base  is  an  equi- 
lateral triangle  with  side  12  ft.,  and  whose  top  is  a line  12  ft.  long 
parallel  to  one  side  of  the  base,  and  whose  altitude  is  15  ft. 

Ex.  30.  If  the  base  of  a prismatoid  is  a rectangle  with  dimensions 
a and  b,  the  top  is  a line  c parallel  to  the  side  b of  the  base,  and  the 
altitude  is  h,  find  the  volume. 


EXERCISES.  CROUP  79 

VOLUMES  OF  CONES  AND  CYLINDERS 

Ex.  1.  How  many  barrels  of  oil  are  contained  in  a cylindrical 
tank  20  ft.  long  and  6 ft.  in  diameter,  if  a barrel  contains  4 eu.  ft.  ? 

Ex.  2.  How  many  eu.  yds.  of  earth  must  be  removed  in  making  a 
tunnel  450  ft.  long,  if  a cross-section  of  the  tunnel  is  a semicircle  of 
15  ft.  radius  ? 

Ex.  3.  A cylindrical  glass  3 in.  in  diameter  holds  half  a pint. 
Find  its  height  in  inches. 

Ex.  4.  If  a cubic  foot  of  brass  be  drawn  out  into  wire  yo  inch  in 
diameter,  how  long  will  the  wire  be  ? 

Ex.  5.  A gallon  measure  is  a cylinder  whose  altitude  equals  the 
diameter  of  the  base.  Find  the  altitude. 

Ex.  6.  Show  that  the  volumes  of  two  cylinders,  having  the  altitude 
of  each  equal  to  the  radius  of  the  other,  are  to  each  other  as  E : E'. 

Ex.  7.  In  a cylinder,  find  E in  terms  of  V and  H ; also  V in  terms 
of  S and  E. 

Ex.  8.  A conical  heap  of  potatoes  is  44  ft.  in  circumference  and 
6 ft.  high.  How  many  bushels  does  it  contain,  if  a bushel  is  lieu,  ft.? 

Ex.  9.  What  fraction  of  a pint  will  a conical  wine-glass  hold,  if 
its  altitude  is  3 in.  and  the  diameter  of  the  top  is  2 in.  ? 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  477 


Ex.  10.  Find  the  ratio  of  the  volumes  of  the  two  cones  inscribed 
in,  and  circumscribed  about,  a regular  tetrahedron. 

Ex.  11.  If  an  equilateral  cone  contains  1 quart,  find  its  dimen- 
sions in  inches. 

Ex.  12.  In  a cone  of  revolution  find  V in  terms  of  R and  L ; also 
find  V in  terms  of  R and  S. 

Ex.  13.  Find  the  volume  of  a frustum  of  a cone  of  revolution, 
whose  radii  are  14  and  7 ft.,  and  whose  altitude  is  2 yds. 

Ex.  14.  What  is  the  cost,  at  50  cts.  a cu.  ft.,  of  a piece  of  marble 
in  the  shape  of  a frustum  of  a cone  of  revolution,  whose  radii  are  6 
and  9 ft.,  and  whose  slant  height  is  5 ft.? 

Ex.  15.  In  a frustum  of  a cone  of  revolution,  the  volume  is  88 
cu.  ft.,  the  altitude  is  9 ft.,  and  R=2r.  Find  r. 


EXERCISES.  CROUP  80 

SPHERICAL  VOLUMES 

Ex.  1.  Find  the  volume  of  a sphere  whose  radius  is  1 ft.  9 in. 

Ex.  2.  If  the  earth  is  a sphere  7,920  miles  in  diameter,  find  its 
volume  in  cubic  miles. 

Ex.  3.  Find  the  diameter  cf  a sphere  whose  volume  is  1 cu.  ft. 

Ex.  4,  What  is  the  volume  of  a sphere  whose  surface  is  616  sq.  in.  ? 

Ex.  5.  Find  the  radius  of  a sphere  equivalent  to  the  sum  of  two 
spheres,  whose  radii  are  2 and  4 in. 

Ex.  6.  Find  the  radius  of  a sphere  whose  volume  equals  the  area 
of  its  surface. 

Ex.  7.  Find  the  volume  of  a sphere  circumscribed  about  a cube 
whose  edge  is  6. 

Ex.  8.  Find  the  volume  of  a spherical  shell  whose  inner  and 
outer  diameters  are  14  and  21  in. 

Ex.  9.  Find  the  volume  of  a spherical  shell  whose  inner  and  outer 
surfaces  are  20  rr  and  12  w. 


478 


SOLID  GEOMETRY 


Find  the  volume  of 

Ex.  10.  A spherical  wedge  whose  angle  is  24°,  the  radius  of  the 
sphere  being  10  in. 

Ex.  11.  A spherical  sector  whose  base  is  a zone  2 in.  high,  the 
radius  of  the  sphere  being  10  in. 

Ex.  12.  A spherical  segment  of  two  bases  whose  radii  are  4 and 
7 and  altitude  5 in. 

Ex.  13.  A wash-basin  in  the  shape  of  a segment  of  a sphere  is 
6 in.  deep  and  24  in.  in  diameter.  How  many  quarts  of  water  will  the 
basin  hold  ? 

Ex.  14.  A plane  parallel  to  the  base  of  a hemisphere  and  bisect- 
ing the  altitude  divides  its  volume  in  what  ratio  ? 

Ex.  15.  A spherical  segment  4 in.  high  contains  200  cu.  in. ; find 
the  radius  of  the  sphere. 

Ex.  16.  If  a heavy  sphere  whose  diameter  is  4 in.  be  placed  in  a 
conical  wine-glass  full  of  water,  whose  diameter  is  5 in.  and  altitude 

6 in.,  find  how  much  water  will  run  over. 

EXERCISES.  CROUP  81 

EQUIVALENT  SOLIDS 

Ex.  1.  If  a cubical  block  of  putty,  each  edge  of  which  is  8 inches, 
be  molded  into  a cylinder  of  revolution  whose  radius  is  3 inches,  find 
the  altitude  of  the  cylinder. 

Ex.  2.  Find  the  radius  of  a sphere  equivalent  to  a cube  whose 
edge  is  10  in. 

Ex.  3.  Find  the  radius  of  a sphere  equivalent  to  a cone  of  revolu- 
tion, whose  radius  is  3 in.  and  altitude  6 in. 

Ex.  4.  Find  the  edge  of  a cube  equivalent  to  a frustum  of  a cone 
of  revolution,  whose  radii  are  4 and  9 ft.  and  altitude  2 yds. 

Ex.  5.  Find  the  altitude  of  a rectangular  parallelop’ped,  whose 
base  is  3 x 5 in.  and  whose  volume  is  equivalent  to  a sphere  of  radius 

7 in. 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  479 


Ex.  6.  Find  the  base  of  a square  rectangular  parallelepiped, 
■whose  altitude  is  8 in.  and  ■whose  volume  equals  the  volume  of  a cone 
of  revolution  with  a radius  of  6 and  an  altitude  of  12  in. 

Ex.  7.  Find  the  radius  of  a cone  of  revolution,  whose  altitude  is 
15  and  whose  volume  is  equal  to  that  of  a cylinder  of  revolution  with 
radius  6 and  altitude  20. 

Ex.  8.  Find  the  altitude  of  a cone  of  revolution,  whose  radius  is 
' 15  and  whose  volume  equals  the  volume  of  a cone  of  revolution  with 
radius  9 and  altitude  24. 

Ex.  9.  On  a sphere  whose  diameter  is  14  the  altitude  of  a zone  of 
one  base  is  2.  Find  the  altitude  of  a cylinder  of  revolution,  whose 
base  equals  the  base  of  the  zone  and  whose  lateral  surface  equals  the 
surface  of  the  zone. 


Ex.  1.  If  on  two  similar  solids  L,  L'  and  l,  V are  pairs  of  homol- 
ogous lines;  A,  A'  and  a,  a'  pairs  of  homologous  areas,  V,  V'  and 
•e,  v'  pairs  of  homologous  volumes, 


Ex.  2.  If  the  edge  of  a cube  is  10  in.,  find  the  edge  of  a cube 
having  5 times  the  surface. 

Ex.  3.  If  the  radius  of  a sphere  is  10  in.,  find  the  radius  of  a 
sphere  having  5 times  the  surface. 

Ex.  4.  If  the  altitude  of  a cone  of  revolution  is  10  in.,  find  the 
altitude  of  a similar  cone  of  revolution  having  5 times  the  surface. 

Ex.  5.  In  the  last  three  exercises,  find  the  required  dimension  if 
the  volume  is  to  be  5 times  the  volume  of  the  original  solid. 

Ex.  6.  The  linear  dimensions  of  one  trunk  are  twice  as  great  as 
those  of  another  trunk.  How  much  greater  is  the  volume  ? The  surface? 


EXERCISES.  CROUP  *2 


SIMILAR  SOLIDS 


480 


SOLID  GEOMETRY 


Ex.  7.  How  far  from  the  vertex  is  the  cross-section  which  bisects 
the  volume  of  a cone  of  revolution  ? Which  bisects  the  lateral 
surface  ? 

Ex.  8.  If  the  altitude  of  a pyramid  is  bisected  by  a plane  parallel 
to  the  base,  how  does  the  area  of  the  cross-section  compare  with  the 
area  of  the  base  f How  does  the  volume  cut  off  compare  with  the 
volume  of  the  entire  pyramid  ? 

Ex.  9.  Planes  parallel  to  the  base  of  a cone  divide  the  altitude 
into  three  equal  parts ; compare  the  lateral  surfaces  cut  off.  Also  the 
volumes. 

Ex.  10.  A sphere  10  in.  in  diameter  is  divided  into  three  equiva- 
lent parts  by  concentric  spherical  surfaces.  Find  the  diameters  of 
these  surfaces. 

Ex.  11.  If  the  strength  of  a muscle  is  as  the  area  of  its  cross- 
section,  and  Goliath  of  Gath  was  three  times  as  large  in  each  linear 
dimension  as  Tom  Thumb,  how  much  greater  was  his  strength  ? His 
weight  ? How,  then,  does  the  activity  of  the  one  man  compare  with 
that,  of  the  other  ? 

Ex.  12.  If  the  rate  at  which  heat  radiates  from  a body  is  in  pro- 
portion to  the  amount  of  surface,  and  the  planet  Jupiter  has  a diame- 
ter 11  times  that  of  the  earth,  how  many  times  longer  will  Jupiter  be 
in  cooling  off  ? 

[Sug.  How  many  times  greater  is  the  volume,  and  therefore  the 
original  amount  of  heat  in  Jupiter  ? How  many  times  greater  is  its 
surface  ? What  will  be  the  combined  effect  of  these  factors  ?] 


CROUP  83 

MISCELLANEOUS  NUMERICAL  EXERCISES  IN  SOLID 
GEOMETRY 

Find  S,  T and  V of 

Ex.  1.  A right  triangular  prism  whose  altitude  is  1 ft.,  and  the 
sides  of  whose  base  are  26,  28,  30  in. 

Ex.  2.  A cone  of  revolution  the  radius  of  whose  base  is  1 ft.  2 in., 
and  whose  altitude  is  35  in. 

Ex.  3.  A frustum  of  a square  pyramid  the  areas  of  whose  bases 
are  1 sq.  ft.  and  36  sq.  in.,  and  whose  altitude  is  9 in. 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  481 


Ex.  4.  A pyramid  whose  slant  height  is  10  in.,  and  whose  base 
is  an  equilateral  triangle  whose  side  is  8 in. 

Ex.  5.  A cube  whose  diagonal  is  1 yd. 

Ex.  6.  A frustum  of  a cone  of  revolution  whose  radii  are  6 and 
11  in.  and  slant  height  13  in. 

Ex.  7.  A rectangular  parallelopiped  whose  diagonal  is  2i/29,  and 
whose  dimensions  are  in  the  ratio  2:3:4. 

Ex.  8.  Find  the  volume  of  a sphere  inscribed  in  a cube  whose 
edge  is  6;  also  find  the  area  of  a triangle  on  that  sphere  whose  angles 
are  80°,  90°,  150°. 

Ex.  9.  Find  the  volume  of  the  spherical  pyramid  whose  base  is 
the  above  triangle. 

Ex.  10.  Find  the  angle  of  a lune  on  the  same  sphere,  equivalent 
to  that  triangle. 

Ex.  11.  On  a cube  whose  edge  is  4,  planes  through  the  midpoints 
of  the  edges  cut  oil  the  corners.  Find  the  volume  of  the  solid  re- 
maining. 

Ex.  12.  How  is  V changed  if  3 of  a cone  of  revolution  is  doubled 
and  R remains  unchanged  ? If  R is  doubled  and  3 remains  unchanged? 
If  both  3 and  R are  doubled  ? 

Ex.  13.  In  an  equilateral  cone,  find  S and  V in  terms  of  R. 

Ex.  14.  A piece  of  lead  20  x8  x2  in.  will  make  how  many  spher- 
ical bullets,  each  f in.  in  diameter  ? 

Ex.  15.  How  many  bricks  are  necessary  to  make  a chimney  in 
the  shape  of  a frustum  of  a cone,  whose  altitude  is  90  ft.,  whose  outer 
diameters  are  3 and  8 ft.,  and  whose  inner  diameters  are  2 and  4 ft., 
counting  12  bricks  to  the  cubic  ft.? 

Ex.  1 6.  If  the  area  of  a zone  is  300  and  its  altitude  6,  find  the 
radius  of  the  sphere. 

Ex.  17.  If  the  section  of  a cylinder  of  revolution  through  its  axis 
is  a square,  find  S,  T,  V in  terms  of  R. 

Ex.  18.  If  every  edge  of  a square  pyramid  is  find  b in  terms  of  T. 


DD 


4S2 


SOLID  GEOMETRY 


Ex.  19.  A regular  square  pyramid  has  a for  its  altitude  and  also  tor 
each  side  of  the  base.  Find  the  area  of  a section  made  by  a plane  parallel 
to  the  base  and  bisecting  the  altitude.  Find  also  the  volumes  of  the 
two  parts  into  which  the  pyramid  is  divided. 

Ex.  20.  If  the  earth  is  a sphere  of  8,000  miles  diameter  and  its 
, atmosphere  extends  50  miles  from  the  earth,  find  the  volume  of  the 
atmosphere. 

Ex.  21.  On  a sphere,  find  the  ratio  of  the  area  of  an  equilateral 
spherical  triangle,  each  of  whose  angles  is  95°,  to  the  area  of  a lune 
whose  angle  is  80°. 

Ex.  22.  A square  right  prism  has  an  altitude  6a  and  an  edge  of 
the  base  2a.  Find  the  volume  of  the  largest  cylinder,  sphere,  pyramid 
and  cone  which  can  be  cut  from  it. 

Ex.  23.  Obtain  a formula  for  the  area  of  that  part  of  a sphere 
illuminated  by  a point  of  light  at  a distance  a from  the  sphere 
whose  radius  is  B. 

Ex.  24.  On  a sphere  whose  radius  is  6 in.,  find  an  angle  of  an 
equilateral  triangle  whose  area  is  12  sq.  in. 

Ex.  25.  Find  the  volume  of  a prismatoid,  whose  altitude  is  24  and 
whose  bases  are  equilateral  triangles,  each  side  10,  so  placed  that  the 
mid-section  of  the  prismatoid  is  a regular  hexagon. 

Ex.  26.  On  a sphere  whose  radius  is  16,  the  bases  of  a zone  are 
equal  and  are  together  equal  to  the  area  of  the  zone.  Find  the  alti- 
tude of  the  zone. 

Ex.  27.  Find  the  volume  of  a square  pyramid,  the  edge  of  whose 
base  is  10  and  each  of  whose  lateral  edges  is  inclined  60°  to  the  base. 

Ex.  28.  An  irregular  piece  of  ore,  if  placed  in  a cylinder  partly 
filled  with  water,  causes  the  water  to  rise  6 in.  If  the  radius  of  the 
cylinder  is  8 in.,  what  is  the  volume  of  the  ore  ? 

Ex.  29.  Find  the  volume  of  a truncated  right  triangular  prism,  if 
the  edges  of  the  base  are  8,  9,  11,  and  the  lateral  edges  are  12,  13,  14. 

Ex.  30.  In  a sphere  whose  radius  is  5,  a section  is  taken  at  the 
distance  3 from  the  center.  On  this  section  as  a base  a cone  i_  formed 
whose  lateral  elements  are  tangent  to  the  sphere.  Find  the  lateral 
surface  and  volume  of  the  cone. 


NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY  483 


Ex,  31.  The  volume  of  a sphere  is  l,437i  eu.  in.  Find  the  surface. 

Ex.  32.  A square  whose  side  is  6 is  revolved  about  a diagonal  as 
an  axis;  find  the  surface  and  volume  generated. 

Ex.  33.  Find  the  edge  of  a cubical  cistern  that  will  hold  10  tons 
of  water,  if  1 cu.  ft.  of  water  weighs  62.28  lbs. 

Ex.  34.  A water  trough  has  equilateral  triangles,  each  side  3 ft., 
for  ends,  and  is  18  ft.  long.  How  many  buckets  of  water  will  it  hold, 
if  a bucket  is  a cylinder  1 ft.  in  diameter  and  11  ft.  high  ? 

Ex.  35.  The  lateral  area  of  a cylinder  of  revolution  is  440  sq.  in., 
and  the  volume  is  1,540  cu.  in.  Find  the  radius  and  altitude. 

Ex.  36.  The  angles  of  a spherical  quadrilateral  are  80°,  100°, 
120°,  120°.  Find  the  angle  of  an  equivalent  equilateral  triangle. 

Ex.  37.  A cone  and  a cylinder  have  equal  lateral  surfaces,  and 
their  axis  sections  are  equilateral.  Find  the  ratio  of  their  volumes. 

Ex.  38.  A water-pipe  4 in.  m diameter  rises  13  ft.  from  the  ground. 
How  many  quarts  of  water  must  be  drawn  from  it  before  the  water 
from  under  the  ground  comes  out  1 If  a quart  runs  out  in  5 seconds, 
how  long  must  the  water  run  ? 

Ex.  39.  A cube  immersed  in  a cylinder  partly  filled  with  water 
causes  the  water  to  rise  4 in.  If  the  radius  of  the  cylinder  is  6 in., 
what  is  an  edge  of  the  cube  ? 

Ex.  40.  An  auger  hole  whose  diameter  is  3 in.  is  bored  through 
the  center  of  a sphere  whose  diameter  is  8 inches.  Find  the  volume 
remaining. 

Ex.  41.  Show  that  the  volumes  of  a cone,  hemisphere,  and  cylin- 
der of  the  same  base  and  altitude  are  as  1 : 2 : 3. 

Ex.  42.  The  volumes  of  two  similar  cylinders  of  revolution  are 
as  8 : 125;  find  the  ratio  of  their  radii.  If  the  radius  of  the  smaller  is 
10  in.,  what  is  the  radius  of  the  larger? 

Ex.  43.  An  iron  shell  is  2 in.  thick  and  the  diameter  of  its  outer 
surface  is  28  in.  Find  its  volume. 

Ex.  44.  The  legs  of  an  isosceles  spherical  triangle  each  make  an 
angle  of  75°  with  the  base.  The  legs  produced  form  a lune  whose 
area  is  four  times  the  area  of  the  triangle.  Find  the  angle  of  the  lune. 


484 


SOLID  GEOMETRY 


CROUP  84 

EXERCISES  INVOLVING  THE  METRIC  SYSTEM 
Find  S,  T,  V of 

Ex.  1.  A right  prism  the  edges  of  whose  base  are  6 m.,  70  dm., 
900  cm.,  and  whose  altitude  is  90  dm. 

Ex.  2.  A regular  square  pyramid  an  edge  of  whose  base  is  30  dm., 
and  whose  altitude  is  1.7  m. 

Ex.  3.  A sphere  whose  radius  is  0.02  m. 

Ex.  4.  A frustum  of  a cone  of  revolution  whose  radii  are  10  dm 
and  6 dm.,  and  whose  slant  height  is  50  cm. 

Ex.  5.  A cube  whose  diagonal  is  12  cm. 

Ex.  6.  A cylinder  of  revolution  whose  radius  equals  2 dm.,  and 
whose  altitude  equals  the  diameter  of  the  base. 

Ex.  7.  Find  the  area  of  a spherical  triangle  on  a sphere  whose 
radius  is  0.02  m.,  if  its  angles  are  110°,  120°,  130°. 

Ex.  8.  Find  the  number  of  square  meters  in  the  surface  of  a 
sphere,  a great  circle  of  which  is  50  dm.  long. 

Ex.  9.  How  many  liters  will  a cylindrical  vessel  hold  that  ig 
10  dm.  in  diameter  and  0.25  m.  high  1 How  many  liquid  quarts  ? 

Ex.  10.  A liter  measure  is  a cylinder  whose  diameter  is  half  the 
altitude.  Find  its  dimensions  in  centimeters. 

Ex.  11.  Find  the  surface  of  a sphere  whose  volume  is  1 cu.  m,, 


APPENDIX 

I.  MODERN  GEOMETRIC  CONCEPTS 


840.  Modern  Geometry.  In  recent  times  many  new 
geometric  ideas  have  been  invented,  and  some  of  them 
developed  into  important  new  branches  of  geometry. 
Thus,  the  idea  of  symmetry  (see  Art.  484,  etc.)  is  a 
modern  geometric  concept.  A few  other  of  these  modern 
concepts  and  methods  will  be  briefly  mentioned,  but 
their  thorough  consideration  lies  beyond  the  scope  of 
this  book. 

841.  Projective  Geometry.  The  idea  of  projections 
(see  Art.  345)  has  been  developed  in  comparatively 
recent  times  into  an  important  branch  of  mathematics  with 
many  practical  applications,  as  in  engineering,  architec- 
ture, construction  of  maps,  etc. 

842.  Principle  of  Continuity.  By  this  principle  two 
or  more  theorems  are  made  special  cases  of  a single  more 
general  theorem.  An  important  aid  in  obtaining  continuity 
among  geometric  principles  is  the  application  of  the  con- 
cept of  negative  quantity  to  geometric  magnitudes.  • 

Thus,  a negative  line  is  a line  opposite  in  direction  to 
a given  line  taken  as  positive. 


For  example,  if  OA  is  H-,  OB  is  — . 

(485) 


4 


48G 


GEOMETEY.  APPENDIX 


Similarly,  a negative  angle  is 
an  angle  formed  by  rotating  a line 
in  a plane  in  a direction  opposite 
from  a direction  of  rotation  taken 
as  positive.  Thus,  if  the  line  OA 
rotating  from  the  position  OA 
forms  the  positive  angle  AOB,  the 

same  line  rotating  in  the  opposite  direction  foiuns  the 
negative  angle  AOB'.  Similarly,  positive  and  negative 
arcs  are  formed. 


In  like  manner,  if  P and  P'  are 
on  opposite  sides  of  the  line  AB  and 
the  area  PAB  is  taken  as  positive, 
the  area  P' AB  will  be  negative. 


As  an  illustration  of  the  law  of 
continuity,  we  may  take  the  theorem 
that  the  sum  of  the  triangles  formed 
by  drawing  lines  from  a point  to  the 
vertices  of  a polygon  equals  the  area 
of  the  polygon. 

Applying  this  to  the  quadrilateral  ABCD,  if  the  point 
P falls  within  the  quadrilateral,  APAP  + APPO  + APCD 
+ A PAD  = ABCB  (Ax.  6). 

Also,  if  the  point  falls  without  the  quadrilateral  at  P', 
A PALP  + A P'BO  + A P'CD  + A PALP  = ABCD , since 
A P AD  is  a negative  area,  and  hence  is  to  be  subtracted 
from  the  sum  of  the  other  three  triangles. 


843.  The  Principle  of  Reciprocity,  or  Duality,  is  a 

principle  of  relation  between  two  theorems  by  which  each 
theorem  is  convertible  into  the  other  by  causing  the  words 
for  the  same  two  geometric  objects  in  each  theorem  to  ex- 
change places. 


MODERN  GEOMETRIC  CONCEPTS 


487 


Thus,  of  theorems  VI  and  VII,  Book  I,  either  may  be 
converted  into  the  other  by  replacing  the  word  "sides”  by 
"angles,”  and  "angles”  by  "sides.”  Hence  these  are 
termed  reciprocal  theorems. 

The  following  are  other  instances  of  reciprocal  geometric 
properties : • 


1.  Two  points  determine  a 
straight  line. 

2.  Three  points  not  in  the  same 
straigh  t line  determine  a plane. 

3.  A straight  line  and  a point 
determine  o plane. 


1.  Two  lines  determine  a point. 

2.  Three  planes  not  through  the 
same  straight  line  determine  a 
point. 

3.  A straight  line  and  a plane 
determine  a point. 


The  reciprocal  of  a theorem  is  not  necessarily  true. 

Thus,  two  parallel  straight  lines  determine  a plane,  but 
two  parallel  planes  do  not  determine  a line. 

However,  by  the  use  of  the  principle  of  reciprocity, 
geometrical  properties,  not  otherwise  obvious,  are  fre- 
quently suggested. 


844.  Principle  of  Homology.  Just  as  the  law  of  reci- 
procity indicates  relations  between  one  set  of  geometric 
concepts  (as  lines)  and  another  set  of  geometric  concepts 
(as  points),  so  the  law  of  homology  indicates  l-elations 
between  a set  of  geometric  concepts  and  a set  of  concepts 
outside  of  geometry:  as  a set  of  algebraic  concepts,  for 
instance. 

Thus,  if  a and  b are  numbers,  by  algebra  (a  + b)  ( a — b) 

9 7,2 

^ — Cl  u • 

Also,  if  a and  b are  segments  of  a line,  the  rectangle 
(«— h6)  X (a — b)  is  equivalent  to  the  difference  between  the 
squares  a?  and  b2. 

By  means  of  this  principle,  truths  which  would  be  over- 
looked or  difficult  to  prove  in  one  department  of  thought 


488 


GEOMETRY.  APPENDIX 


are  made  obvious  by  observing  the  corresponding  truth  in 
another  department  of  thought. 

Thus,  if  a and  b are  line  segments,  the  theorem  (a-\-  b )2 
“Ha — 6)2  = 2(a2  + &2)  is  not  immediately  obvious  in  geo- 
metry, but  becomes  so  by  observing  the  like  relation 
between  the  algebraic  numbers  a and  b. 

845.  Non-Euclidean  Geometry.  Hyperspace.  By  vary- 
ing the  properties  of  space,  as  these  are  ordinarily  stated, 
different  kinds  of  space  may  be  conceived  of,  each  having 
its  own  geometric  laws  and  properties.  Thus,  space,  as 
we  ordinarily  conceive  it,  has  three  dimensions,  but  it  is 
possible  to  conceive  of  space  as  having  four  or  more 
dimensions.  To  mention  a single  property  of  four  dimen- 
sional space,  in  such  a space  it  would  be  possible,  by 
simple  pressure,  to  turn  a sphere,  as  an  orange,  inside 
out  without  breaking  its  surface. 

As  an  aid  toward  conceiving  how  this  is  possible,  consider  a plane 
in  which  one  circle  lies  inside  another.  No  matter  how  these  circles 
are  moved  about  in  the  plane,  it  is  impossible  to  shift  the  inner  circle 
so  as  to  place  it  outside  the  other  without  breaking  the  circumference 
of  the  outer  circle.  But,  if  we  are  allowed  to  use  the  third  dimension 
of  space,  it  is  a simple  matter  to  lift  the  inner  circle  up  out  of  the 
plane  and  set  it  down  outside  the  larger  circle. 

Similarly  if,  in  space  of  three  dimensions,  we  have  one  spherical 
shell  inside  a larger  shell,  it  is  impossible  to  place  the  smaller  shell 
outside  the  larger  without  breaking  the  larger.  But  if  the  use  of  a 
fourth  dimension  be  allowed,— that  is,  the  use  of  another  dimension 
of  freedom  of  motion, — it  is  possible  to  place  the  inner  shell  outside 
the  larger  without  breaking  the  latter. 

846.  Curved  Spaces.  By  varying  the  geometric  axioms 
of  space  (see  Art.  47),  different  kinds  of  space  may  be 
conceived  of.  Thus,  we  may  conceive  of  space  such  that 
through  a given  point  one  line  may  be  drawn  parallel  to  a 
given  line  (that  is  ordinary,  or  Euclidean  space);  or  such 


MODERN  GEOMETRIC  CONCEPTS 


489 


that  through  a given  point  no  line  can  be  drawn  parallel 
to  a given  line  (spherical  space);  or  such  that  through  a 
given  point  more  than  one  line  can  be  drawn  parallel  to  a 
given  line  (pseudo -spherical  space). 

These  different  kinds  of  space  differ  in  many  of  their 
properties.  For  example,  in  the  first  of  them  the  sum  of 
the  angles  of  a triangle  equals  two  right  angles;  in  the 
second,  it  is  greater;  in  the* third,  it  is  less. 

These  different  kinds  of  space,  however,  have  many 
properties  in  common.  Thus,  in  all  of  them  every  point  in 
the  perpendicular  bisector  of  a line  is  equidistant  from  the 
extremities  of  the  line. 

EXERCISES.  CROUP  86 

Ex.  1.  Show  by  the  use  of  zero  and  negative  ares  that  the  princi- 
ples of  Arts.  257,  263,  258,  264,  265,  are  particular  cases  of  the  general 
theorem  that  the  angle  included  between  two  lines  which  cut  or  touch 
a circle  is  measured  by  one-half  the  sunf  of  the  intercepted  arcs. 

Ex.  2.  Show  that  the  principles  of  Arts.  354  and  358  are  particular 
cases  of  the  theorem  that,  if  two  lines  are  drawn  from  or  through  a 
point  to  meet  a circumference,  the  product  of  the  segments  of  one  line 
equals  the  product  of  the  segments  of  the  other  line. 

Ex.  3.  Show  by  the  use  of  negative  angles 
that  theorem  XXXVIII,  Book  I,  is  true  for  a 
quadrilateral  of  the  form  ABCD.  \_BCD  is  a 
negative  angle;  the  angle  at  the  vertex  D is 
the  reflex  angle  ADC.] 

Ex.  4.  What  is  the  reciprocal  of  the  state- 
ment that  two  intersecting  straight  lines  deter  ■ 
mine  a plane  ? 

Ex.  5.  What  is  the  reciprocal  of  the  statement  that  three  planes 
perpendicular  to  each  other  determine  three  straight  lines  perpen- 
dicular to  each  other  ? 


n.  HISTORY  OF  GEOMETRY 


847.  Origin  of  Geometry  as  a Science.  The  beginnings 

<of  geometry  as  a science  are  found  in  Egypt,  dating  back 
at  least  three  thousand  years  before  Christ.  Herodotus 
says  that  geometry,  as  knowfl  in  Egypt,  grew  out  of  the 
need  of  remeasuring  pieces  of  land  parts  of  which  had  been 
washed  away  by  the  Nile  floods,  in  order  to  make  an  equi- 
table readjustment  of  the  taxes  on  the  same. 

The  substance  of  the  Egyptian  geometry  is  found  in  an 
old  papyrus  roll,  now  in  the  British  museum.  This  roll 
is,  in  effect,  a mathematical  treatise  written  by  a scribe 
named  Ahmes  at  least  1700  B.C.,  and  is,  the  writer  states, 
a copy  of  a more  ancient  work,  dating,  say,  3000  B.  C. 

848.  Epochs  in  the.  Development  of  Geometry.  From 
Egypt  a knowledge  of  geometry  was  transferred  to  Greece, 
whence  it  spread  to  other  countries.  Hence  we  have  the 
following  principal  epochs  in  the  development  of  geometry: 

1.  Egyptian  : 3000  B.  C.— 1500  B.  C. 

2.  Greek  : 600  B.  C.— 100  B.  C. 

3.  Hindoo  : 500  A.  D— 1100  A.  D. 

4.  Arab  : 800  A.  D.— 1200  A.  D. 

5.  European  : 1200  A.  D. 

In  the  year  1120  A.  D , Athelard,  an  English  monk, 
visited  Cordova,  in  Spam,  in  the  disguise  of  a Mohamme- 
dan student,  and  procured  a copy  of  Euclid  m the  Arabic 
language.  This  book  he  brought  back  to  central  Europe, 
where  it  was  translated  into  Latin  and  became  the  basis  of 
all  geometric  study  in  Europe  till  the  year  1533,  when, 

(490) 


HISTORY  OF  GEOMETRY 


491 


owing  to  the  capture  of  Constantinople  by  the  Turks, 
copies  of  the  works  of  the  Greek  mathematicians  in  the 
original  Greek  were  scattered  through  Europe. 


HISTORY  OF  GEOMETRICAL  METHODS 

849.  Rhetorical  Methods.  By  rhetorical  methods  in 
the  presentation  of  geometric  truths,  is  meant  the  use  of 
definitions,  axioms,  theorems,  geometric  figures,  the  rep- 
resentation of  geometric  magnitudes  by  the  use  of  letters, 
the  arrangement  of  material  in  Books,  etc.  The  Egyptians 
had  none  of  these,  their  geometric  knowledge  being  re- 
corded only  in  the  shape  of  the  solutions  of  certain  numeri- 
cal examples,  from  which  the  rules  used  must  be  inferred. 

Thales  (Greece  GOO  B.C.)  first  made  an  enunciation  of 
an  abstract  property  of  a geometric  figure.  He  had  a rude 
idea  of  the  geometric  theorem. 

Pythagoras  (Italy  525,  B.C.)  introduced  formal  defini- 
tions into  geometry,  though  some  of  those  used  by  him 
were  not  very  accurate.  For  instance,  his  definition  of  a 
point  is  "unity  having  position.”  Pythagoras  also 
arranged  the  leading  propositions  known  to  him  in 
something  like  logical  order. 

Hippocrates  (Athens,  420  B.C.)  was  the  first  systemati- 
cally to  denote  a point  by  a capital  letter,  and  a segment 
of  a line  by  two  capital  letters,  as  the  line  AB,  as  is  done 
at  present.  He  also  wrote  the  first  text -book  on  geometry. 

Plato  (Athens,  380  B.  C.)  made  definitions,  axioms 
and  postulates  the  beginning  and  basis  of  geometry. 

To  Euclid  (Alexandria,  280  B.  C.)  is  due  the  division 
of  geometry  into  Books,  the  formal  enunciation  of  theo- 
rems, the  particular  enunciation,  the  formal  construction, 


492 


GEOMETRY.  APPENDIX 


proof,  and  conclusion,  in  presenting  a proposition.  Hs 
also  introduced  the  use  of  the  corollary  and  scholium. 

Using  these  methods  of  presenting  geometric  truths. 
Euclid  wrote  a text -book  of  geometry  in  thirteen  books 
■which  was  the  standard  text-book  on  this  subject  for 
nearly  two  thousand  years. 

The  use  of  the  symbols  A,  CO  , ||  , etc.,  in  geometric 
proofs  originated  in  the  United  States  in  recent  years. 

850.  Logical  Methods.  The  Egyptians  used  no  formal 
methods  of  proof.  They  probably  obtained  their  few  crude 
geometric  processes  as  the  result  of  experiment. 

The  Hindoos  also  used  no  formal  proof.  One  of  their 
writers  on  geometry  merely  states  a theorem,  draws  a 
figure,  and  says  "Behold  ! ” 

The  use  of  logical  methods  of  geometric  proof  is  due  to 
the  Greeks.  The  early  Greek  geometricians  used  experi- 
mental methods  at  times,  in  order  to  obtain  geometric 
truths.  For  instance,  they  determined  that  the  angles  at 
the  base  of  an  isosceles  triangle  are  equal,  by  folding  half 
of  the  triangle  over  on  the  altitude  as  an  axis  and  observ- 
ing that  the  angles  mentioned  coincided  as  a fact,  but 
without  showing  that  they  must  coincide. 

Pythagoras  (525  B.  C.)  was  the  first  to  establish  geo- 
metric truths  by  systematic  deduction , but  his  methods 
wei’e  sometimes  faulty.  For  instance,  he  believed  that  the 
converse  of  a proposition  is  necessarily  true. 

Hippocrates  (420  B.  C.)  used  correct  and  rigorous  de- 
duction in  geometric  proofs.  He  also  introduced  specific 
varieties  of  such  deduction,  such  as  the  method  of  reduc- 
ing one  proposition  to  another  (Art.  296),  and  the  reductio 
ad  absurdum. 


HISTORY  OF  GEOMETRY 


493 


The  methods  of  deduction  used  by  the  Greeks,  however,  were  de- 
fective in  their  lack  of  generality.  For  instance,  it  was  often  thought 
necessary  to  have  a separate  proof  of  a theorem  for  each  different 
kind  of  figure  to  which  the  theorem  applied. 

Thus,  the  theorem  that  the  sum  of  the  an- 
gles of  a triangle  equals  two  right  angles 
was  proved, 

(1)  for  the  equilateral  triangle  by  use 
of  the  regular  hexagon ; 

(2)  for  the  right  triangle  by  the  use  of 
a rectangle ; 

(3)  for  a scalene  triangle  by  dividing 
the  scalene  triangle  into  two  right  triangles. 

The  Greeks  appeared  to  fear  that  a 
general  proof  might  be  vitiated  if  it  were 
applied  to  a figure  in  any  way  special  or 
peculiar.  In  other  words,  they  had  no  conception  of  the  principle  of 
continuity  (Art.  842). 

Plato  (380  ‘B.  C.)  introduced  the  method  of  proof  by 
analysis , that  is,  by  taking  a proposition  as  true  and  work- 
ing from  it  back  to  known  truths  (see  Art.  196). 

To  Eudoxus  (360  B.  C.)  is  virtually  due  proof  by  the 
method  of  limits ; though  his  method,  known  as  the 
method  of  exhaustions,  is  crude  aud  cumbersome. 

Apollonius  (Alexandria,  225  B.  C.)  used  projections , 
transversals , etc.,  which.,  in  modern  times,  have  developed 
into  the  subject  of  projective  geometry. 

851.  Mechanical  Methods.  The  Greeks,  in  demonstra- 
ting a geometrical  theorem,  usually  drew  the  figure  em- 
ployed in  a bed  of  sand.  This  method  had  certain  advan- 
tages, but  was  not  adapted  to  the  use  of  a large  audience. 

At  the  time  when  geometry  was  being  developed  in  Greece,  the 
interest  in  the  subject  was  very  general.  There  was  scarcely  a town 
but  had  its  lectures  oa  the  subject.  The  news  of  the  discovery  of  a 


494 


GEOMETRY.  APPENDIX 


new  theorem  spread  from  town  to  town,  and  the  theorem  was  redemon- 
strated in  the  sand  of  each  market  place. 

The  Greek  treatises,  however,  were  written  on  vellum 
or  papyrus  by  the  use  of  the  reed,  or  calamus,  and  ink. 

In  Roman  times,  and  in  the  middle  ages,  geometrical 
figures  were  drawn  in  wax  smeared  on  wooden  boards, 
called  tablets.  They  were  drawn  by  the  use  of  the  stylus, 
a metal  stick,  pointed  at  one  end  for  making  marks,  and 
broad  at  the  other  for  erasing  marks.  These  wax  tablets 
were  still  in  use  in  Shakespeare’s  time  (see  Hamlet  Act  I, 
Sc.  5,  1.  107).  The  blackboard  and  crayon  are  modem 
inventions,  their  use  having  developed  within  the  last  one 
hundred  years. 

The  Greeks  invented  many  kinds  of  drawing  instruments 
for  tracing  various  curves.  It  was  due  to  the  influence  of 
Plato  (380  B.  C.)  that,  in  constructing  geometric  figures, 
the  use  of  only  the  ruler  and  compasses  is  permitted. 


HISTORY  OF  GEOMETRIC  TRUTHS.  PLANE  GEOMETRY 


852.  Rectilinear  Figures.  The  Egyp- 
tians measured  the  area  of  any  four- 
sided field  by  multiplying  half  the  sum 
of  one  pair  of  opposite  sides  by  half  the 
sum  of  the  other  pair;  which  was  equivalent  to  using  the 

a-\-c  b-\-d 

formula,  area  -X  0 • 


This,  of  course,  gives  a correct  result  for  the  rectangle  and  square, 
but  gives  too  great  a result  for  other  quadrilaterals,  as  the  trapezoid, 
etc.  Hence  Joseph,  of  the  Book  of  Genesis,  in  buying  the  fields  of 
the  Egyptians  for  Pharoah  in  time  of  famine  by  the  use  of  this 
formula,  in  many  cases  paid  for  a larger  field  than  he  obtained. 


The  Egyptians  had  a special  fondness  for  geometrical 
constructions,  probably  growing  out  of  their  work  as  temple 


HISTORY  OF  GEOMETRY 


495 


builders.  A class  of  workers  existed  among  them  called 
"rope -stretchers,”  whose  business  was  the  marking  out  of 
the  foundations  of  buildings.  These  men  knew  how  to 
bisect  an  angle  and  also  to  construct  a right  angle.  The 
latter  was  probably  done  by  a method  essentially  the  same 
as  forming  a right  triangle  whose  sides  are  three,  four  and 
five  units  of  length.  Ahmes,  in  his  treatise,  has  various 
constructions  of  the  isosceles  trapezoid  from  different  data. 

Thales  (COO  B.  C.)  enunciated  the  following  theorems: 

If  two  straight  lines  intersect,  the  opposite  or  vertical 
angles  are  equal; 

The  angles  at  the  base  of  an  isosceles  triangle  are  equal; 

Two  triangles  are  equal  if  two  sides  and  the  included 
angle  of  one  are  equal  to  two  sides  and  the  included  angle 
of  the  other; 

The  sum  of  the  angles  of  a triangle  equals  two  right 
angles; 

Two  mutually  equiangular  triangles  are  similar. 

Thales  used  the  last  of  these  theorems  to  measure  tha 
height  of  the  great  pyramid  by  measuring  the  length  of 
the  shadow  cast  by  the  pyramid  and  also  measuring  the 
length  of  the  shadow  of  a post  of  known  height  at  the  same 
time  and  making  a proportion  between  these  quantities. 

Pythagoras  (525  B.  C.)  and  his  followers  discovered 
correct  formulas  for  the  areas  of  the  principal  rectilinear 
figures,  and  also  discovered  the  theorems  that  the  areas  of 
similar  polygons  are  as  the  squares  of  their  homologous 
sides,  and  that  the  square  on  the  hypotenuse  of  a right 
triangle  equals  the  sum  of  the  squares  on  the  other  two 
sides.  The  latter  is  called  the  Pythagorean  theorem. 
They  also  discovered  how  to  construct  a square  equivalent 
to  a given  parallelogi’am,  and  to  divide  a given  line  in 
mean  and  extreme  ratio, 


496 


GEOMETRY.  APPENDIX 


To  Eudoxus  (380  B.  C.)  we  owe  the  general  theory  of 
proportion  in  geometry,  and  the  treatment  of  incommen- 
surable quantities  by  the  method  of  Exhaustions.  By  the 
use  of  these  he  obtained  such  theorems  as  that  the  areas 
of  two  circles  are  to  each  other  as  the  squares  of  their 
radii,  or  of  their  diameters. 

In  the  writings  of  Hero  (Alexandria,  125  B.  C.)  we  first 
find  the  formula  for  the  area  of  a triangle  in  terms  of  its 
sides,  K=V/ s(s — a)  ( s — b)  ( s — c).  Hero  also  was  the  first 
to  place  land-surveying  on  a scientific  basis. 


It  is  a curious  fact  that  Hero  at  the  same  time  gives  an  incorrect 
formula  for  the  area  of  a triangle,  viz.,  K=ia(b-\-c) , this  formula 
being  apparently  derived  from  Egyptian  sources. 


Xenodorus  (150  B.  C.)  investigated  isoperemetrical 
figures. 


The  Romans,  though  they  excelled  in  engineering,  ap- 
parently did  not  appreciate  the  value  of  the  Greek  geom- 
etry. Even  after  they  became  acquainted  with  it,  they 
continued  to  use  antiquated  and  inaccurate  formulas  for 
areas,  some  of  them  of  obscure  origin.  Thus,  they  used 
the  Egyptian  formula  for  the  area  of  a quadrilateral, 

2T=— X— ; w-.  They  determined  the  area  of  an  equilat- 

Li 

eral  triangle  whose  side  is  a,  by  different  formulas,  all 


incorrect,  as  K— 


13ff2 
30  ’ 


K=i(o?-\-a) , and  K—ia2. 


853.  The  Circle.  Thales  enunciated  the  theorem  that 
every  diameter  bisects  a circle,  and  proved  the  theorem 
that  an  angle  inscribed  in  a semicircle  is  a right  angle. 

To  Hippocrates  (420  B.  C.)  is  due  the  discovery  of 
nearly  all  the  other  principal  properties  of  the  circle  given 
in  this  book. 


HISTORY  OF  GEOMETRY 


497 


The  Egyptians  regarded  the  area  of  the  circle  as  equiva- 
lent to  of  the  diameter  squared,  which  would  make 
*■  = 3.1004. 

The  Jews  and  Babylonians  treated  7t  as  equal  to  3. 

Archimedes,  by  the  use  of  inscribed  and  circumscribed 
regular  polygons,  showed  that  the  true  value  of  ~l  lies 
between  3?  and  3t7;  that  is,  between  3.14285  and  3.1408. 

The  Hindoo  writers  assign  various  values  to  7 1,  as  3,  3i, 
l/10,  and  Aryabhatta  (530  A.  D.)  gives  the  correct  ap- 
proximation, 3. 1410.  The  Hindoos  used  the  formula 

1/2 1/4 (See  Art.  468)  in  computing  the  numeri- 

cal value  of  7t. 

Within  recent  times,  the  value  of  7t  has  been  computed 
to  707  decimal  places. 

The  use  of  the  symbol  71  for  the  ratio  of  the  circum- 
ference of  a circle  to  the  diameter  was  established  in 
mathematics  by  Euler  (Germany,  1750). 

HISTORY  OF  GEOMETRIC  TRUTHS.  SOLID  GEOMETRY 

854.  Polyhedrons.  The  Egyptians  computed  the  vol- 
umes of  solid  figures  from  the  linear  dimensions  of  such 
figures.  Thus,  Alimes  computes  the  contents  of  an  Egyp- 
tian barn  by  methods  which  are  equivalent  to  the  use 

3c 

of  the  formula  V=  aXbX  — . As  the  shape  of  these  barns 

is  not  known,  it  is  not  possible  to  say  whether  this  formula 
is  correct  or  not. 

Pythagoras  discovered,  or  knew,  all  the  regular  poly- 
hedrons except  the  dodecahedron.  These  polyhedrons  were 
supposed  to  have  various  magical  or  mystical  properties. 
Hence  the  study  of  them  was  made  very  prominent. 

FF 


498 


GEOMETRY.  APPENDIX 


Hippasus  (470  B.  C.)  discovered  the  dodecahedron,  but 
he  was  drowned  by  the  other  Pythagoreans  for  boasting 
of  the  discovery. 

Eudoxus  (380  B.  C.)  showed  that  the  volume  of  a pyra- 
mid is  equivalent  to  one -third  the  product  of  its  base  by 
its  altitude. 

E.  F.  August  (Germany,  1849)  introduced  the  prisma- 
toid  formula  into  geometry  and  showed  its  importance. 

855.  The  Three  Round  Bodies.  Eudoxus  showed  that 
the  volume  of  a cone  is  equivalent  to  one -third  the  area  of 
its  base  by  its  altitude. 

Archimedes  discovered  the  formulas  for  the  surface  and 
volume  of  the  sphere. 

Menelaus  (100  A.  D.)  treated  of  the  properties  of 
spherical  triangles. 

Gerard  (Holland,  1620)  invented  polar  triangles  and 
found  the  formulas  for  the  area  of  a spherical  triangle  and 
of  a spherioal  polygon. 

856.  Non-Euclidean  Geometry.  The  idea  that  a space 
might  exist  having  different  properties  from  those  which 
we  regard  as  belonging  to  the  space  in  which  we  live,  has 
occurred  to  different  thinkers  at  different  times,  but 
Lobatchewsky  (Russia,  1793-1856)  was  the  first  to  make 
systematic  use  of  this  principle.  He  found  that  if,  instead 
of  taking  Geom.  Ax.  2 as  true,  we  suppose  that  through  a 
given  point  in  a plane  several  straight  lines  may  be  drawn 
parallel  to  a given  line,  the  result  is  not  a series  of  absur- 
dities or  a general  reductio  ad  absurdum;  but,  on  the  con- 
trary, a consistent  series  of  theorems  is  obtained  giving 
the  properties  of  a space. 


III.  REVIEW  EXERCISES 

EXERCISES.  CROUP  86 

"review  exercises  in  plane  geometry 

Ex.  1.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular 
to  each  other,  the  angles  are  supplementary. 

Ex.  2.  If  a diagonal  of  a quadrilateral  bisects  two  of  its  angles, 
the  diagonal  bisects  the  quadrilateral. 

Ex.  3.  Through  a given  point  draw  a secant  at  a given  distance 
from  the  center  of  a given  circle. 

Ex.  4.  The  bisector  of  one  angle  of  a triangle  and  of  an  exterior 
angle  at  another  vertex  form  an  angle  which  is  equal  to  one-half  the 
third  angle  of  the  triangle. 

Ex.  5.  The  side  of  a square  is  18  in.  Find  the  circumference  of 
the  inscribed  and  circumscribed  circles. 

Ex.  6.  The  quadrilateral  ADBC  is  inscribed  in  a circle.  The  diag- 
onals AB  and  DC  intersect  in  the  point  F.  Arc  AD  = 112°,  arc  AC  = 
108°,  lAFC  = 74°.  Find  all  the  other  angles  of  the  figure. 

Ex.  7.  Find  the  locus  of  the  center  of  a circle  which  touches  two 
given  equal  circles. 

Ex.  8.  Find  the  area  of  a triangle  whose  sides  are  1 m.,  17  dm., 
210  cm. 

Ex.  9.  The  line  joining  the  midpoints  of  two  radii  is  perpendicular 
to  the  line  bisecting  their  angle. 

Ex.  10.  If  a quadrilateral  be  inscribed  in  a circle  and  its  diag- 
onals drawn,  how  many  pairs  of  similar  triangles  are  formed  ? 

Ex.  11.  Prove  that  the  sum  of  the  exterior  angles  of  a polygon 
(Art.  172)  equals  four  right  angles,  by  the  use  of  a figure  formed  by 
drawing  lines  from  a point  within  a polygon  to  the  vertices  of  the 
polygon. 


500 


GEOMETRY.  APPENDIX 


Ex.  12.  In  a circle  whose  radius  is  12  cm.,  find  the  length  of  the 
tangent  drawn  from  a point  at  a distance  240  mm.  from  the  center. 

Ex.  13.  If  two  sides  of  a regular  pentagon  he  produced,  find  the 
angle  of  their  intersection. 

Ex.  14.  In  the  parallelogram  ABCD,  points  are  taken  on  the 
diagonals  such  that  AP=BQ=CB  = DS.  Show  that  PQES  is  a 
parall  elogram. 

Ex.  15.  A chord  6 in.  long  is  at  the  distance  4 in.  from  the  center 
of  a circle.  Find  the  distance  from  the  center  of  a chord  8 in.  long. 

Ex.  16.  If  B is  a point  in  the  circumference  of  a circle  whose 
center  is  O,  PA  a tangent  at  any  point  P,  meeting  OB  produced  at  At 
and  PD  perpendicular  to  OB,  then  PB  bisects  the  angle  APD. 

Ex.  17.  Construct  a parallelogram,  given  a side,  an  angle,  and  ai 
diagonal. 

Ex.  18.  Find  in  inches  the  sides  of  an  isosceles  light  triangle 
whose  area  is  1 sq.  yd. 

Ex.  19.  Given  the  line  a,  construct  ° 1 ~ — — . 

Ex.  20.  If  two  lines  intersect  so  that  the  product  of  the  segments 
of  one  line  equals  the  product  of  the  segments  of  the  other,  a cir- 
cumference may  be  passed  through  the  extremities  of  the  two  lines. 

Ex.  21.  Find  the  locus  of  the  vertices  of  all  triangles  on  a given 
base  and  having  a given  area. 

Ex.  22.  On  the  figure  p.  206,  prove  that  BCi-\-AFa=AB2-\-FC2- 

Ex.  23.  The  area  of  a rectangle  is  108  and  the  base  is  three  times 
tii-.  altitude.  Find  the  dimensions. 

Ex.  24.  If,  on  the  sides  AC  and  BC  of  the  triangle  ABC,  the 
squares,  AD  and  BF,  are  constructed,  AF  and  DB  are  equal. 

Ex.  25.  If  the  angle  included  between  a tangent  and  a secant  is 
half  a right  angle,  and  the  tangent  equals  the  rad'us,  the  secant 
passes  through  the  center  of  the  circle. 


REVIEW  EXERCISES  IN  PLANE  GEOMETRY  501 


Ex.  26.  The  sum  of  the  areas  of  two  circles  is  20  sq.  yds.,  and  the 
difference  of  their  areas  is  15  sq.  yds.  Find  their  radii. 

Ex.  27.  Construct  an  isosceles  trapezoid,  given  the  bases  and  a 
leg. 

Ex.  28.  Show  that,  if  the  alternate  sides  of  a regular  pentagon 
be  produced  to  meet,  the  points  of  intersection  formed  are  the  vertices 
of  another  regular  pentagon. 

Ex.  29.  If  a post  2 ft.  G in.  high  casts  a shadow  1 ft.  9 in.  long, 
how  tall  is  a tree  which,  at  the  same  time,  casts  a shadow  60  ft.  long  1 

Ex.  30.  If  two  intersecting  chords  make  equal  angles  with  the 
diameter  through  their  point  of  intersection,  the  chords  are  equal. 

Ex.  31.  From  a given  point  draw  a secant  to  a circle  so  that  the 
external  segment  is  half  the  secant. 

Ex.  32.  Find  the  locus  of  the  center  of  a circle  which  touches  a 
given  circle  at  a given  point. 

Ex.  33.  If  one  diagonal  ot  a quadrilateral  bisects  the  other 
diagonal,  the  first  diagonal  divides  the  quadrilateral  into  two  equi- 
valent triangles. 

Ex.  34.  In  a given  square  inscribe  a square  having  a given  side. 

Ex.  35.  A field  in  the  shape  of  an  equilateral  triangle  contains 
one  acre.  How  many  feet  does  one  side  contain  ? 

Ex.  36.  If  perpendiculars  are  drawn  to  a given  line  from  the  ver- 
tices of  a parallelogram,  the  sum  of  the  perpendiculars  from  two 
opposite  vertices  equals  the  sum  of  the  other  two  perpendiculars. 

Ex.  37.  Any  two  altitudes  of  a triangle  are  reciprocally  propor 
lional  to  the  bases  on  which  they  stand. 

Ex.  38.  Construct  a triangle  equivalent  to  a given  triangle  and 
having  two  given  sides. 

Ex.  39.  The  apothem  of  a regular  hexagon  is  20.  Find  the  area 
of  the  inscribed  and  circumscribed  circles. 

Ex.  40.  M is  the  midpoint  of  the  hypotenuse  AB  of  a right  tri- 
angle ABC.  Prove  8 jZc2=Zb2+RC2-(-MC2. 


502 


GEOMETRY.  APPENDIX 


Ex.  41.  Transform  a given  triangle  into  an  equivalent  right  tri- 
angle containing  a given  acute  angle. 

Ex.  42.  The  area  of  a square  inscribed  in  a semicircle  is  to  the 
area  of  the  square  inscribed  in  the  circle  as  2 : 5. 

Ex.  43.  If,  on  a diameter  of  the  circle  0,  OA=OB  and  AC  is  par- 
allel to  BD,  the  chord  CD  is  perpendicular  to  AC. 

Ex.  44.  Find  the  radius  of  a circle  whose  area  is  equal  to  one- 
third  the  area  of  the  circle  whose  radius  is  7 in. 

Ex.  45.  State  and  prove  the  converse  of  Prop.  XXI,  Book  III. 

Ex.  46.  If,  in  a given  trapezoid,  one  base  is  three  times  the  other 
base,  the  segments  of  each  diagonal  are  as  1 : 3. 

Ex.  47.  If  two  sides  of  a triangle  are  6 and  12  and  the  angle 
included  by  them  is  60°,  had  the  length  of  the  other  side.  Also  find 
this  when  the  included  angle  is  45°;  also,  when  120°. 

Ex.  48.  How  many  sides  has  a polygon  in  which  the  sum  of  the 
interior  angles  exceeds  the  sum  of  the  exterior  angles  by  540°? 

Ex.  49.  If  the  four  sides  of  a quadrilateral  are  the  diameter  of  a 
circle,  the  two  tangents  at  its  extremities,  and  a tangent  at  any  other 
point,  the  area  of  the  quadrilateral  equals  one-half  the  product  of  the 
diameter  by  the  side  opposite  it  in  the  quadrilateral. 

Ex.  50.  An  equilateral  triangle  and  a regular  hexagon  have  the 
same  perimeter;  find  the  ratio  of  their  areas. 

Ex.  51.  To  a circle  whose  radius  is  30  cm.  a tangent  is  drawn 
from  a point  21  dm.  from  the  center.  Find  the  length  of  the  tangent. 

Ex.  52.  If  two  opposite  sides  of  a quadrilateral  are  equal,  and 
the  angles  which  they  make  with  a third  side  are  equal,  the  quad- 
rilateral is  a trapezoid. 

Ex.  53.  If  two  circles  are  tangent  externally  and  two  parallel 
diameters  are  drawn,  one  in  each  circle,  a pair  of  opposite  extremities 
of  the  two  diameters  and  the  point  of  contact  are  eollinear. 

Ex.  54.  If,  in  the  triangle  ABC,  the  line  AD  is  perpendicular  to 
BD,  the  bisector  of  the  angle  B,  a line  through  D parallel  to  PC 
bisects  AC. 


REVIEW  EXERCISES  IN  PLANE  GEOMETRY  503 


Ex.  55.  Bisect  a given  triangle  by  a line  parallel  to  a given  line. 

Ex.  56.  If  two  parallelograms  have  an  angle  of  one  equal  to  the 
supplement  of  an  angle  of  the  other,  their  areas  are  to  each  other  as 
the  products  of  the  sides  including  the  angles. 

Ex.  57.  The  sum  of  the  medians  of  a triangle  is  less  than  the 
perimeter,  and  greater  than  half  the  perimeter. 

Ex.  58.  If  PARE  is  a secant  to  a circle  through  the  center  O,  PT 
a tangent,  and  TR  perpendicular  to  PB,  then  PA  : PR=PO  : PB. 

Ex.  59.  Two  concentric  circles  have  radii  of  17  and  15.  Find  the 
'ength  of  the  chord  of  the  larger  which  is  tangent  to  the  smaller. 

Ex.  60.  On  the  figure,  p.  244, 

(а)  Find  two  pairs  of  similar  triangles; 

(б)  Find  two  dotted  lines  which  are  perpendicular  to  each  other; 

(c)  Discover  a theorem  concerning  points,  not  connected  by  lines 
on  the  figure,  which  are  collinear; 

(d)  Discover  a theorem  concerning  squares  on  given  lines. 

Ex.  61.  One  of  the  legs,  AC,  of  an  isosceles  triangle  is  produced 
through  the  vertex,  C,  to  the  point  F,  and  F is  joined  with  D,  the  mid- 
point of  the  base  AB.  DF  intersects  BC  in  E.  Prove  that  CF  is 
greater  than  CE. 

Ex.  62.  The  line  of  centers  of  two  circles  intersects  their  common 
external  tangent  at  P.  PABCD  is  a secant  intersecting  one  of  the 
two  circles  at  A and  B and  the  other  at  C and  D.  Prove  PAX  PE= 
PBXPC. 

Ex.  63.  Trisect  a given  parallelogram  by  lines  drawn  through  a 
given  vertex. 

Ex.  64.  Find  the  area  of  a triangle  the  sides  of  which  are  the 
chord  of  an  arc  of  120°  in  a circle  whose  radius  is  1 ; the  chord  of  an 
arc  of  90°  in  a circle  whose  radius  is  2 ; and  the  chord  of  an  are  of 
80°  in  a circle  whose  radius  is  3. 

Ex.  65.  Construct  a triangle,  given  the  median  to  one  side  and  a 
median  and  altitude  on  the  other  side. 

Ex.  66.  Two  circles  intersect  at  P and  Q.  The  chord  CQ  is  tan- 
gent to  the  circle  QPB  at  Q.  APB  is  any  chord  through  P.  Prove 
that  AC' is  parallel  to  BQ. 


504 


GEOMETRY.  APPENDIX 


Ex.  67.  In  the  triangle  ABC,  from  D,  the  midpoint  of  BC,  BE 
and  DF  are  drawn,  bisecting  the  angles  ADB  and  ADC,  and  meeting 
AB  at  E and  AC  at  F.  Prove  EF  ||  BC. 

Ex.  68.  Produce  the  side  BC  of  the  triangle  ABC  to  a point  P,  sc 
that  PBXPC=PA2. 

Ex.  69.  In  a given  circle  inscribe  a rectangle  similar  to  a given 
rectangle. 

Ex.  70.  In  a given  semicircle  inscribe  a rectangle  similar  to  S 
given  rectangle. 

Ex.  71.  The  area  of  an  isosceles  trapezoid  is  140  sq.  ft.,  one  base 
is  26  ft.,  and  the  legs  make  an  angle  of  45°  with  the  other  base.  Find 
the  other  base. 

Ex.  72.  Cut  oil  one-third  the  area  of  a given  triangle  by  a line 
perpendicular  to  one  side. 

Ex.  73.  Find  the  sides  of  a triangle  whose  area  is  1 sq.  ft.,  if 
the  sides  are  in  the  ratio  2:3:4. 

Ex.  74.  Divide  a given  line  into  two  parts  such  that  the  sum  of 
the  squares  of  the  two  parts  shall  be  a minimum. 

Ex.  75.  If,  from  any  point  in  the  base  of  a triangle,  lines  are 
drawn  parallel  to  the  sides,  find  the  locus  of  the  center  of  the  paral- 
lelogram so  formed. 

Ex.  76.  Three  sides  of  a quadrilateral  are  845,  613,  810,  and  the 
fourth  side  is  perpendicular  to  the  sides  845  and  810.  Find  the  area. 

Ex.  77.  If  BP  bisects  the  angle  ABC,  and  DP 
bisects  the  angle  CD  A,  prove  that  angle  P=i  sum 
of  angles  A and  C. 

Ex.  78.  Two  circles  intersect  at  P and  Q. 

Through  a point  A in  one  circumference  lines  APC 
and  AQD  are  drawn,  meeting  xhe  other  in  C and  D.  Prove  the  tan- 
gent at  A parallel  to  CD. 

Ex.  79.  In  a given  triangle,  draw  a line  parallel  to  the  base  and 
terminated  by  the  sides  so  that  it  shall  be  a mean  proportional  be* 
tween  the  segments  of  one  sidt 


REVIEW  EXERCISES  IN  PLANE  GEOMETRY  505 


Ex.  80.  Find  the  angle  inscribed  in  a semicircle  the  sum  of  whose 
Bides  is  a maximum. 

Ex.  81.  The  bases  of  a trapezoid  are  160  and  120,  and  the  alti- 
tude 140.  Find  the  dimensions  of  two  equivalent  trapezoids  into 
which  the  given  trapezoid  is  divided  by  a line  parallel  to  the  base. 

Ex.  82.  If  the  diameter  of  a given  circle  be  divided  into  any  two 
segments,  and  a semicircumferenee  be  described  on  the  two  segments 
on  opposite  sides  of  the  diameter,  the  area  of  the  circle  will  be  di- 
vided by  the  semicireumferences  thus  drawn  into  two  parts  having 
the  same  ratio  as  tlm  segments  of  the  diameter. 

Ex.  83.  On  a given  straight  line,  AB,  two  segments  of  circles  are 
drawn,  APB  and  AQB.  The  angles  QAP  and  QBP  are  bisected  by  lines 
meeting  in  B.  Prove  that  the  angle  If  is  a constant,  wherever  P and 
Q may  be  on  their  ares. 

Ex.  84.  On  the  side  AB  of  the  triangle  ABC,  as  diameter,  a cir- 
cle is  described.  EF  is  a diameter  parallel  to  BC.  Show  that  EB 
bisects  tho  angle  ABC. 

Ex.  85.  Construct  a trapezoid,  given  the  bases,  one  diagonal,  and 
an  angle  included  by  the  diagonals. 

Ex.  86.  If,  through  any  point  in  the  common  chord  of  two  inter- 
secting circles,  two  chords  be  drawn,  one  in  each  circle,  through  the 
four  extremities  of  the  two  chords  a circumference  may  be  passed. 

Ex.  87.  From  a given  point  as  center  describe  a circle  cutting  a 
given  straight  line  in  two  points,  so  that  the  product  of  the  distances 
of  the  points  from  a given  point  in  the  line  may  equal  the  square  of  a 
given  line  segment. 

Ex.  88.  AB  is  any  chord  in  a given  circle,  P any  point  on  the 
circumference,  PM  is  perpendicular  to  AB  and  is  produced  to  meet 
the  circle  at  Q;  AN  is  drawn  perpendicular  to  the  tangent  at  P. 
Prove  the  triangles  NAM  and  PAQ  similar. 

Ex.  89.  If  two  circles  ABCD  and  EBCF  intersect  in  B and  C and 
have  common  exterior  tangents  AE  and  DF  cut  by  BC  produced  at  G- 
and  B,  then  GM2=BC2+AE2, 


506 


GEOMETEY.  APPENDIX 


EXERCISES.  CROUP  87 

REVIEW  EXERCISES  IN  SOLID  GEOMETRY 

Ex.  1.  A segment  of  a straight  line  oblique  to  a plane  is  greater 
than  its  projection  on  the  plane. 

Ex.  2.  Two  tetrahedrons  are  similar  if  a dihedral  angle  of  one 
equals  a dihedral  angle  of  the  other,  and  the  faces  forming  these 
dihedral  angles  are  similar  each  to  each. 

Ex.  3.  A plane  and  a straight  line,  both  of  which  are  parallel  to 
the  same  line,  are  parallel  to  each  other. 

Ex.  4.  If  the  diagonal  of  one  face  of  a cube  is  10  inches,  find  the 
volume  of  the  cube. 

Ex.  5.  Construct  a spherical  triangle  on  a give 
poles  of  the  sides  of  the  triangle. 

Ex.  6.  Given  AB  ± M, 

AE  and  BF  _L  MB ; 
prove  EF  X PM. 

Ex.  7.  The  diagonals  of  a rectangular  paral- 
lelopiped  are  equal. 

Ex.  8.  What  portion  of  the  surface  of  a sphere 
is  a triangle  each  of  whose  angles  is  140° f 

Ex.  9.  Through  a given  point  pass  a plane  parallel  to  two  given 
straight  lines. 

Ex.  10.  Show  that  the  lateral  area  of  a cylinder  of  revolution  is 
equivalent  to  a circle  whose  radius  is  a mean  proportional  between 
the  altitude  of  the  cylinder  and  the  diameter  of  its  base. 

Ex.  11.  The  volumes  of  polyhedrons  circumscribed  about  equal 
spheres  are  to  each  other  as  the  surfaces  of  the  polyhedrons. 

Ex.  12.  Find  S and  T of  a regular  square  pyramid  an  edge  of 
whose  base  is  14  dm.,  and  whose  lateral  edge  is  250  cm. 

Ex.  13.  If  two  lines  are  parallel  and  a plane  be  passed  through  each 
line,  the  intersection  of  these  planes  is  parallel  to  the  given  lines. 


3n  sphere,  given  the 


REVIEW  EXERCISES  IN  SOLID  GEOMETRY  507 


Ex.  14.  Given  PH  L plane  AD, 
l PEH= Z PFH ; 
prove  Z PEF=  Z PFE. 

Ex.  15.  If  a plane  be  passed 
through  the  midpoints  of  three 
edges  of  a parallelopiped  which 
meet  at  a vertex,  the  pyramid  thus  formed  is  what  part  of  the 
parallelopiped  ? 

Ex.  16.  Find  a point  in  a plane  such  that  the  sum  of  its  distances 
from  two  given  points  on  the  same  side  of  the  plane  is  a minimum. 

Ex.  17.  Given  the  points  A,  B,  C,  D in  a plane  and  P a point 
outside  the  plane,  AB  perpendicular  to  the  plane  PBD,  and  AC  per- 
pendicular to  the  plane  PCD;  prove  that  PD  is  perpendicular  to 
the  plane  ABCD. 

Ex.  18.  In  a sphere  whoso  radius  is  5,  find  the  area  of  a zone  the 
radii  of  whose  upper  and  lower  bases  are  3 and  4. 

Ex.  19.  Two  cylinders  of  revolution  have  equal  lateral  areas. 
Show  that  their  volumes  are  as  B : R' . 

Ex.  20.  The  midpoints  of  two  opposite  sides  of  a quadrilateral  in 
space,  and  the  midpoints  of  its  diagonals,  are  the  vertices  of  a 
parallelogram. 

Ex.  21.  How  many  feet  of  two-inch  plank  are  necessary  to  con- 
struct a box  twice  as  wide  as  deep  and  twice  as  long  as  wide  (on  the 
inside),  and  to  contain  216  cu.  ft.? 

Ex.  22.  If  two  spheres  with  radii  R and  r are  concentric,  find  the 
area  of  the  section  of  the  larger  sphere  made  by  a plane  tangent  to 
the  smaller  sphere. 

Ex.  23.  In  the  frustum  of  a regular  square  pyramid,  the  edges 
of  the  bases  are  denoted  by  b i and  b>  and  the  altitude  by  H;  prove 
that  i = il/(&1— 62)2  + 4HL 

Ex.  24.  If  the  opposite  sides  of  a spherical  quadrilateral  are  equal 
the  opposite  angles  are  equal. 


508 


GEOMETRY.  APPENDIX 


Ex.  25.  Obtain  the  simplest  formula  for  the  lateral  surface  of  a 
truncated  triangular  right  prism,  each  edge  of  whose  base  is  a,  and 
whose  lateral  edges  are  p,  q,  and  r. 

Ex.  26.  The  area  of  a zone  of  one  base  is  a mean  proportional 
between  the  remaining  surface  of  the  sphere  and  its  entire  surraee. 
Find  the  altitude  of  the  zone. 

Ex.  27.  The  lateral  edges  of  two  similar  frusta  are  as  1 : a.  How 
do  their  areas  compare  ? Their  volumes  f 

Ex.  28.  Construct  a spherical  surface  with  a given  radius,  r,  which 
shall  be  tangent  to  a given  plane,  and  to  a given  sphere,  and  also  pass 
through  a given  point. 

Tx.  29.  The  volume  of  a right  circular  cylinder  equals  the  area 
of  the  generating  rectangle  multiplied  by  the  circumference  generated 
by  the  point  of  intersection  of  its  diagonals. 

Ex.  30.  On  a sphere  whose  radius  is  8$  inches,  find  the  area  of  a 
zone  generated  by  a pair  of  compasses  whose  points  are  5 inches 
apart. 

Ex.  31.  The  perpendicular  to  a given  plane  from  the  point  where 
the  altitudes  of  a regular  tetrahedron  intersect  equals  one-fourth  the 
sum  of  the  perpendiculars  from  the  vertices  of  the  tetrahedron  to  the 
same  plane. 

Ex.  32.  Two  trihedral  angles  are  equal  or  symmetrical  if  their 
corresponding  dihedral  angles  are  equal. 

Ex.  33.  On  a sphere  whose  radius  is  a,  a zone  has  equal  bases 
and  the  sum  of  the  bases  equals  the  area  of  the  zone.  Find  the  alti- 
tude of  the  zone. 

Ex.  34.  A plane  which  bisects  two  opposite  edges  of  a tetrahedron 
bisects  the  volume  of  the  tetrahedron. 

Ex.  35.  Find  the  locus  of  all  points  in  space  which  have  their 
distances  from  two  given  parallel  lines  in  a given  ratio. 

Ex.  36.  If  a,  b,  c are  the  sides  of  a spherical  triangle,  a',  cf 
the  sides  of  its  polar  triangle,  and  a>b>c,  then  a'<V<c! . 

Ex.  37.  A cone  of  revolution  has  a lateral  area  of  4 sq.  yd.  and 
an  altitude  of  2 ft.  How  much  of  the  altitude  must  be  cut  off  by  a 
plane  parallel  to  the  base,  in  order  to  leave  a frustum  whose  lateral 
area  is  2 sq,  ft.  ? 


REVIEW  EXERCISES  IN  SOLID  GEOMETRY  509 


Ex.  38.  The  total  area  of  an  equilateral  cone  is  to  the  area  of  the 
inscribed  sphere  as  9 : 4. 

Ex.  39.  Construct  a sphere  of  given  radius,  r,  whose  surface  shall 
be  tangent  to  three  given  spheres. 

Ex.  40.  The  volume  of  the  frustum  of  an  equilateral  cone  is  300 
eu.  in.  and  its  altitude  is  20  in.  Show  how  to  find  the  radii  of  the  bases. 

Ex.  41.  On  each  base  of  a cylinder  of  revolution  a cone  is  placed, 
with  its  vertex  at  the  center  of  the  opposite  base.  Find  the  radius  of 
the  circle  of  intersection  of  the  two  conical  surfaces. 

Ex.  42.  The  volume  of  a frustum  of  a cone  of  revolution  equals 
the  sum  of  a cylinder  and  a cone  of  the  same  altitude  as  the  frustum, 
and  with  radii  which  are  respectively  the  half  sum  and  the  half  differ- 
ence of  the  radii  of  the  frustum. 

Ex.  43.  A square  whose  side  is  a revolves  about  a line  through 
one  of  its  vertices  and  parallel  to  a diagonal,  as  axis;  find  the  surface 
and  volume  generated. 

Ex.  44.  If  a cone  of  revolution  roll  on  another  fixed  cone  of  revo- 
lution so  that  their  vertices  coincide,  find  the  kind  of  surface  gen- 
erated by  the  axis  of  the  rolling  cone. 

Ex.  45.  An  equilateral  triangle  whose  side  is  a revolves  about  an 
altitude  as  an  axis;  find  the  surface  and  volume  generated  by  the 
inscribed  circle,  and  also  by  the  circumscribed  circle. 

Ex.  46.  Find  the  locus  of  the  center  of  a sphere  which  is  tan- 
gent to  three  given  planes. 

Ex.  47.  If  an  equilateral  triangle  whose  side  is  a be  rotated  about 
a line  through  one  vertex  and  parallel  to  the  opposite  side,  as  an  axis, 
find  the  surface  and  volume  generated. 

Ex.  48.  What  other  formulas  of  solid  geometry  may  be  regarded 
as  special  cases  of  the  formula  for  the  volume  of  a prismatoid  ? 

Ex.  49.  Through  a given  point  pass  a plane  which  shall  bisect  the 
volume  of  a given  tetrahedron. 

Ex.  50.  In  an  equilateral  cone  and  a cone  whose  opposite  ele- 
ments are  perpendicular  at  the  vertex,  show  that  the  ratio  of  the 
Vertical  solid  angles  is  as  2 — j/3  : 2 — 1/2. 


PRACTICAL  APPLICATIONS  OF  PLANE  GEOMETRY 


EXERCISES.  GROUP  88 

(Book  I) 

1.  Take  a piece  of  paper  having  a straight  edge  and  fold  the  paper 
so  as  to  form  a right  angle.  What  geometrical  principle  or  definition 
have  you  used? 

2.  By  use  of  a board  with  a straight  edge,  test  the  accuracy  of  the 

■fn  outside  of  a carpenter’s  square  by  a method 

indicated  in  the  diagram.  How,  then,  would 
I you  test  the  accuracy  of  the  inside  angle  of 

1 .1  the  square?  What  geometric  principle  have  you 

used  in  each  case? 

3.  In  Ex.  2 prove  that  the  error  in  the  outside  angle  of  the  carpenter’s 
square,  if  there  be  any,  equals  one-half  the  angle,  x , between  the  out- 
side lines  of  the  square  as  shown  in  the  diagram  (denote  the  error  by  e 
and  show  that  e + e = x). 

This  principle  is  important  because  it  is  essentially  the  method  used 
in  correcting  the  axis  of  a telescope,  and  hence  in  correcting  instru- 
ments of  which  the  telescope  is  a part,  as  various 
surveying  and  astronomical  instruments. 

4.  By  use  of  a carpenter’s  square  and  a given 
straight  edge,  lay  off  a series  of  parallel  lines.  What 
property  of  parallel  lines  have  you  used? 

5.  Tell  how  to  construct  a carpenter’s  miter  box. 

6.  The  diagram  shows  a drawing  instrument  called 
the  parallel  rulers.  The  dotted  outline  shows  the  rest 
of  the  instrument  in  another  position,  the  part  RS 
remaining  fixed.  The  distance  PQ  = RS,  PR  = QS. 

Hence  show  that  PQ  is.  parallel  to  RS. 

In  like  manner  show  that  P'Q'  is  parallel  to  RS. 

Hence  show  PQ  is  parallel  to  P’Q'. 

If  a line  be  drawn  perpendicular  to  PQ  and 
another  perpendicular  to  P'Q’,  the  perpendiculars  thus  drawn  will  be 
parallel  to  each  other  (Art.  122,  lines  perpendicular  to  parallel  lines 
are  parallel). 

510 

Copyright,  1911 

8r  Cbabl-es  E.  MEBBiLb  Compact 


PRACTICAL  APPLICATIONS 


511 


The  above  are  cases  of  linked  motion,  a kind  of  mechanism  of  wide 
importance.  Observe  for  instance  the  system  of  links  which  connect 
the  driving  wheels  of  a locomotive  with  the  piston  in  the  cylinder,  and 
also  the  jointed  rods  connecting  the  walking  beam  of  a steamboat  with 
the  engine.  Look  up  also,  in  the  Century  Dictionary,  for  instance, 
the  words  linkage,  cell,  and  parallel  motion. 

7.  The  distance  between  two  accessible  places  separated  by  an  im- 
passable barrier  (as  between  two  houses  separated  by 
a pond)  may  be  found  by  the  following  method  when 
no  instrument  for  measuring  angles  is  at  hand. 

Let  A and  B be  the  twd  places.  Take  a conven- 
ient station  C and  measure  AC  and  BC.  Produce  AC 
to  F,  making  CF  = AC.  Produce  BC  to  D,  making 
DC  = BC.  Measure  DF.  Prove  AB  = DF. 

If  AC  = CF  = 220  ft.;  BC  = CD  = 190  ft.;  and  DF  = 210  ft.,  how 
long  is  AB? 

8.  In  the  trusses  of  steel  bridges,  why  are  the  beams  and  rods 
arranged  so  as  to  form  a network  of  triangles  as  far  as  possible,  and  not 
of  quadrilaterals,  or  pentagons,  for  instance?  (See  Ex.  3,  p.  76;  also 
Art.  101.) 

How  is  this  principle  also  made  use  of  in  forming  the  frame  of  a wooden 
house,  or  box  car,  or  to  strengthen  a weak  frame  or  fence  of  any  kind? 

9.  Draw  a map  for  the  following  survey  notes  to  the  scale  of 

400  ft.  to  the  inch.  In  laying 
off  the  angles  draw  a dotted 
north  and  south  line  through 
each  station  and  then  use  a 
protractor. 

Keep  your  drawing  for  a 
later  use. 

10.  Obtain  or  make  up  a set 
of  survey  notes  similar  to  those  in  Ex.  9 and  make  a drawing  for  them. 

11.  Let  DC  and  FC  (p.  512)  be  two  walls  perpendicular  to  the  plane 
of  the  paper.  Let  small  mirrors  be  attached  to  these  walls  at  A and  B. 
Let  Z.  C = 45°.  Let  a ray  of  light  pass  through  Q to  A,  be  reflected 
to  B,  and  thence  to  P.  Prove  that  Z APB  is  a right  angle. 

[Sug.  The  law  of  reflected  light  is  that  the  angle  of  incidence  equals 
the  angle  of  reflection,  or,  in  the  figure,  x = x and  y = y.  Then  in 
the  triangle  A B C,  x + y + 45°  = 180°,  or  x + y = 135°.  About  the 
points  A and  B,  2x+a+2y-rb  = 360° a + b = 90°,  etc.] 


Stations 

Bearings 

Distances 

A 

N. 30°  E. 

300  ft. 

B 

N. 

200  ft. 

C 

S.  67°  E. 

450  ft. 

D 

S.  60°  W. 

600  ft. 

512 


GEOMETRY 


'0  . 


The  preceding  is  the  principle  of  an  instrument  called  the  “optical 
square”  used  by  foresters  in  constructing  right  angles.  For  a ray  of 
light  coming  from  R through  a small  hole  at  B 
above  the  mirror  will  make  a right  angle  with  the 
ray  coming  from  Q through  P to  A. 

12.  The  velocity  of  light  is  determined  by  the 
use  of  a rotating  mirror  in  the  following  manner: 
Let  AiBi  be  a mirror  _L  plane  of  the  paper  and 
rotating  about  0 as  a pivot;  let  OPi  be  _|_  AiBi. 
Let  LO  be  a ray  of  light  striking  the  mirror  at  0 and  reflected  through 
M to  a small  stationary  mirror  some  miles  distant,  whence  it  is 
reflected  back  through  M to  0.  On  the  return  of  the  ray  to  0, 
AiBi  will  have  rotated  through  a 
small  angle,  a,  to  the  position  AzBz, 
hence  the  ray  will  be  reflected  in  the 
direction  OR.  Let  the  pupil  show  that 
Z a = \ Z LOR.  Since  Z LOR  may 
readily  be  measured,  Z a.  is  known,  and 
if  the  rate  at  which  the  mirror  is  rotat- 
ing is  known,  the  time  occupied  by  the 
ray  in  traveling  from  0 to  the  stationary  mirror  and  back  is  determined. 

[Sug.  ZPiOP2  = Z A,0A2  (Art.  132) 

ZPiOL  = Z PiOM  = x ( Z incidence  = ^/reflection) 

Z PiOL  = x — a 
Z LOR  — x + a — (x  — a)  = 2 a] 

If  Z LOR  = 2°  19',  OM  = 3 mi.,  and  the  mirror  AB  rotates  100 
times  per  second,  determine  the  velocity  of  light  per  second. 

13.  To  prove  that  the  image  of  a point  in  a plane  mirror  is  on  a 
perpendicular  from  the  point  to  the  mirror  and  as  far  behind  the 

mirror  as  the  object  is  in  front  of  it,  let 
MM'  be  the  mirror  and  P the  point,  and 
PAE  and  PA'E'  two  reflected  rays.  Show 
that 

Z PAM  = z EAM'  = z P' AM. 

:.  Z PAM'  = z P' AM'. 

Hence  A PAA'  = A P'AA'. 

PA  = P'A,  etc. 

14.  From  the  above  show  the  direction  in  which  a billiard  ball  must 
be  sent  in  order  to  strike  a certain  point  on  the  table  after  striking  one 
side  of  the  table 


M 


PRACTICAL  APPLICATIONS 


513 


15.  If  two  mirrors  OM  and  OM'  be  perpendicular  to  each  other 

make  a construction  to  show 
where  the  point  P will  appear 
to  be  to  an  eye  at  E,  after  the 
light  from  P has  been  reflected 
from  both  mirrors. 

[Sug.  Prove  J3P  " = BA  + AP' 
= BA  + AP.] 

16.  What  would  be  the  appli- 
cation of  Ex.  15  to  a ball  struck 
on  a billiard  table? 

17.  The  path  of  a ray  of  light 
before  and  after  entering  a glass 
prism  is  given  by  the  lines  AB 

and  CD.  The  entire  angle  by  which  a 
ray  of  light  is  deflected  on  passing  through 
the  prism  is  denoted  by  x.  Prove  that 
x = i + r — P.  <y_s  PBy  and  PCy  are  rt.  /_s.) 
[Stjg.  By  use  of  a quadrilateral  BPCy, 
^ y = 180°  — P.  Then  use  the  quadrilateral 
all  of  whose  sides  are  dotted  lines.] 


EXERCISES.  GROUP  89 

(Book  II) 

1.  Given  a fragment  of  broken  wheel  show  how  to  find  the  radius 
of  the  wheel. 

2.  By  use  of  the  carpenter’s  square  find  the 
center  of  a given  circle. 

3.  Show  how  a pattern  maker  by  the  use  of  a 
carpenter’s  square  can  determine  whether  the 
cavity  made  in  the  edge  of  a board  or  piece  of 
metal  is  a semicircle. 

4.  Bisect  a given  angle  by  the  use  of  a carpenter’s  square. 

6.  By  use  of  squared  paper  divide  a line  1J  inches  long  into  five 
equal  parts.  Into  7 equal  parts.  Into  3 equal  parts.  Can  you  make 
this  division  on  paper  ruled  in  only  one  direction? 

6.  Make  up  and  work  a similar  example  for  yourself. 


514 


GEOMETRY 


7.  Draw  a line  AB  1|  in.  long  and  let  0 be  its  midpoint.  Markoff 
OC  = \ in.  and  divide  AC  into  four  equal  parts  at  D,  F,  H.  With  0 
as  a center  draw  circumferences  through  C,  D,  F,  H,  A.  From  0 draw 

radii  (one  of  which  is  OA)  dividing  the 
angular  space  about  0 into  eight  equal 
parts.  Beginning  at  0 draw  the 
smooth  curve  C123456  through  the 
points  where  the  radii  drawn  inter- 
sect the  circumferences  as  indicated 
on  the  diagram.  The  result  will  be 
the  outline  of  a special  form  of  wheel 
called  a cam,  much  used  in  machines, 
as  in  the  sewing  and  harvesting 
machines,  printing  presses,  automo- 
biles, weaving  looms,  and  machinery 
for  making  shoes.  A cam  converts  a 
circular  motion  into  a reciprocating  or 
back  and  forth  motion.  The  difference  of  length  between  OB  and  OC 
on  the  diagram  is  termed  the  “throw”  of  the  cam. 

8.  Make  a drawing  of  a cam  in  which  the  throw  shall  be  1 in.  and 
the  longest  radius  If  in. 

9.  Stretch  a 100  ft.  tape  or  part  of  it  so  as  to  construct  an  angle 
of  60°  as  accurately  as  possible. 

10.  To  extend  a straight  line  AB  beyond  an  obstacle,  as  a building, 
we  may  proceed  as  follows: 


B 


C 

A 


H 


D 


At  B measure  off  an  angle 
ABC  = 60°.  Produce  CB  to  D, 
etc.  Let  the  pupil  complete  the  a. — 
construction  and  prove  that  his 
method  is  correct. 

11.  Let  the  pupil  solve  the 
problem  of  Ex.  10  by  the  construction  of  right  angles  instead  of  angles 
of  60°. 

12.  To  find  the  distance  between  two  points  A and  B,  one  of  which 
(B)  is  inaccessible,  we  may  proceed  as  fol- 
lows: Extend  BA  to  C.  Measure  a con- 
venient line  AD  and  extend  AD  to  E, 
making  DE  = AD.  From  E run  a hue  j| 
AC  and  meeting  BD  extended  at  F. 
Measure  EF.  Prove  that  EF  = AB. 


PRACTICAL  APPLICATIONS 


513 


13.  Show  how  to  solve  the  problem  of  Ex.  12  by  constructing  a 
line  at  right  angles  to  another  line  instead  of  one  parallel  to  another. 

14.  If  two  streets  meet,  as  in  the  diagram, 
show  how  a curve  of  given  radius  r may  be 
made  to  take  the  place  of  the  angle  A and  be 
tangent  with  the  curb  of  the  two  streets. 

15.  The  curve  in  a railroad  track  is  usually 
an  arc  of  a circle  tangent  to  each  straight  track 

which  it  joins.  If  two  straight  tracks  AB  and 
CD  are  joined  by  a circular  curve  tangent  to 
both  of  them,  and  P is  the  point  where  AB  and 
CD  would  meet  if  extended,  prove  Z TPD  = 2 
Z TAC. 

16.  One  way  of  laying  off  a railroad  curve 
tangent  to  a given  track  is  as  follows:  Let  ABP  be  the 
given  straight  track.  At  B construct  a small  angle  PBC 
(whose  size  depends  on  the  degree  of  curvature  which 
the  curve  is  to  have).  On  it  mark  off  BC  = 100  ft. 

Construct  Z CBD  = Z PBC.  Take  C as  a center  and 
100  ft.  as  a radius,  and  describe  an  arc  cutting  BD 
Construct  E from  D in  like  manner.  Prove 
that,  B,  C,  D,  E all  he  on 
the  arc  of  a circumference  tangent  to  AB 
at  B. 

[Stjg.  Pass  a circumference  through  B, 
C,  D.  Prove  that  E lies  on  this  circumfer- 
ence and  that  ABP  is  tangent  to  it.] 

17.  Let  AB  and  CD  be  two  straight  rail- 
D road  tracks  connected  by  an  arc  AF  of  a 
circle  whose  center  is  0,  and  an  arc  FC 
whose  center  is  O',  the  arcs  having  a com- 
mon tangent  at  F.  Prove  that  the  angle  of 
intersection  ( x ) of  the  two  straight  tracks,  if  produced,  equals  the  sum 
of  the  central  angles  of  the  two  tracks. 

[Sug.  x = y + z.  Use  Ex.  15. j 

A curve  like  AFC  composed  of  two  or  more  arcs  of  different  radii  is 
called  a compound  curve.  Can  you  suggest  why  a compound  curve 
should  be  used  in  connecting  railroad  tracks? 

18.  If  the  two  arcs  which  compose  a compound  curve  He  on  opposite 
‘tides  of  their  common  tangent,  the  compound  curve  is  called  a reverse 


at  D. 


/ 


GEOMETRY 


516 


curve.  Thus  on  the  diagram,  BCD  is  a reverse  curve  connecting  the 

straight  roads  AB  and  DE. 


the  angles  x,  y,  z. 

[Stjg.  Use  trian- 
gle HFE] 

Reverse  curves  are 
much  used  in  archi- 
tecture and  ornamen- 
tal work. 


* \v 

"'s\  / 

'F 


19.  What  is  a railroad  frog?  If  a curved  track  crosses  a straight 

track,  show  that  the  angle  of  the  frog  ( x ) equals  the  ^ 

central  angle  of  the  curved  track  (o). 

20.  If  two  tracks  which  curve  in  the  same  direction 
cross  each  other,  find  the  relation  between  the  angle 


of  the  frog  and  the  central  angles  of  the  two  curved  tracks. 

21.  Find  the  same  when  the  two  curved  tracks  curve  in  opposite 
directions. 

22.  Show  how  to  locate  a gas-generating  plant  (P)  along  a given 

p straight  road  AB  so  that  the  length  of  pipe 

j connecting  it  with  two  towns  ( C and  L) 

\ j shall  be  a minimum. 

^v-./  (Use  Ex.  24,  p.  176.) 

APB 

23.  Discover  and  state  an  application  of 
Ex.  24,  p.  176,  similar  to  one  given  in  Ex.  22. 

24.  If  a treasure  has  been  buried  100  ft.  from  a certain  tree,  and  equi- 
distant from  a given  straight  road  and  a path  parallel  to  the  road,  show 
how  the  treasure  may  be  found. 

26.  Make  up  and  work  a similar  example  for  yourself  using  the  prin- 
ciple of  Ex.  2,  p.  167.  Also  using  that  of  one  of  Exs.  3-8,  p.  168. 

26.  Prove  that  the  latitude  of  a place  on  the 
earth’s  surface  equals  the  elevation  of  the  pole 
(that  is,  on  the  diagram,  prove  /_  QEA  = Z.PAO). 


27.  Given  the  sun’s  declina- 
tion ( i.e . distance  north  or  south 
of  the  celestial  equator),  show 
how  to  determine  the  latitude 
of  a place  by  measuring  the  zenith  distance  of  the 
sun.  Also  by  measuring  the  altitude  of  the  sun  above  the  horizon. 


PRACTICAL  APPLICATIONS 


517 


28.  How  was  Peary  aided  by  the  principles  of  Exs.  26  and  27  in 
determining  whether  he  had  arrived  at  the  North  Pole? 

29.  If  on  April  6 (the  day  of  the  year  on  which  Peary  was  at  the 
North  Pole)  the  sun  was  6°  7'  north  of  the  celestial  equator,  how  high 
above  the  horizon  should  the  sun  have  been  as  observed  by  Peary? 
At  what  hour  of  the  day  was  this? 

30.  The  sextant  is  used  almost  daily  by  every  navigator  in  deter- 
mining the  altitude  of  the  sun  above  the  horizon,  and  hence  the  latitude 
of  the  ship.  The  construction  of  the  sextant  is 
based  on  the  principle  that  if  a ray  of  light  be 
reflected  from  two  mirrors  in  succession,  the 
change  in  the  direction  of  the  ray  of  light 
equals  twice  the  angle  made  by  the  mirrors. 

Prove  this  law. 

[Sug.  Let  M and  N be  the  mirrors  and  the 
reflected  ray  be  SMNO.  It  is  required  to 
prove  that  Z.y  = 2 Z_x.  It  is  to  be  noted 
that  Z.FNM , being  an  exterior  angle  of  triangle 
NS'M,  equals  x + i.  Hence  angle  MNO  equals 
180°  — 2 (x  + i).  Hence  in  triangle  MNO, 
y + 180°  - 2 (x  + i)  + 2 i = 180°,  etc.] 

In  the  sextant,  angle  y is  the  elevation  of  the 
sun  above  the  horizon,  and  x can  be  read  on  the  rim  of  the  instru- 
g ment  in  a simple  manner,  since  the 

mirror  N is  fixed  in  position  while 
M rotates. 

It  may  be  observed  that  the  prin- 
ciple of  this  example  is  the  same  as 
that  proved  in  Ex.  12,  p.  512. 

31.  The  diagram  shows  a some- 
what simple  substitute  for  the  sex- 
tant called  the  angle-meter.  0 is  the 
center  of  the  arc  BC  and  MM'  is  a 
fixed  mirror.  The  instrument  is 
held  so  that  a ray  SO  from  the  sun 
when  reflected  from  the  mirror  passes 
into  the  eye  (A)  in  a horizontal  direc- 
tion. Show  that  the  elevation  of  the 
sun  above  the  horizon  = 2 /_  x.  The  rim  BC  is  graduated  so  that 
the  angle  x can  be  read  on  it. 


\x, 

V 


518 


GEOMETRY 


The  angle-meter  may  be  used  to  measure  horizontal  as  well  as  vertical 
angles. 

What  is  the  advantage  in  using  a mirror  on  an  instrument  of  this 
sort?  That  is,  why  is  not  the  position  of  the  sun  observed  directly 
across  a graduated  arc? 


32.  Draw  an  easement  cornice  ( AB ) tangent  to  the  rake  cornice  BC, 
\ q and  passing  through  a required  point  A.  (The 

//  same  construction  is  used  in  laying  out  the  ease- 
ments of  stair  rails,  etc.) 

(Use  Ex.  7,  p.  174.) 

33.  A segmental  arch  is  a compound  curve 
composed  of  the  arcs  of  three  circles.  The  method 
of  constructing  a segmental  arch  is  as  follows: 
Let  AB  be  the  span  and  CD  the  altitude  of  the  re- 
quired arch.  Complete  the  rectangle  GADC. 

Draw  the  diagonal  AC.  Bisect  the  angles 
GAC  and  GCA.  Let  the  bisectors  meet  at  E. 

Draw  EH  perpendicular  to  AC,  meeting  AB  at 
N and  CD  produced  at  H.  Make  DK  equal  to 
DN.  Then  show  that  AT  is  the  center  and  NA 
the  radius,  H the  center  and  HE  the  radius, 
and  K the  center  and  KB  the  radius  for  the  arcs  composing  the 
arch.  (See  Hanstein’s  Constructive  Drawing.) 

[Sug.  At  E draw  a line  perpendicular  to  EH.  Then  Z HE  A = 
Z NAE.  (Complements  of  equal  angles  are  equal)  etc.] 

34.  What  is  called  a Persian  arch  may  be  constructed  as  follows: 
Let  AB  be  the  span  and  CD  the  altitude  of  the  required  arch.  Draw 
F B F the  isosceles  triangles  ADB.  Divide  AD  into  three 
equal  parts  at  H and  G.  Construct  HK  _L  AG 
at  H,  and  meeting  AB  produced  at  K.  Produce 
KG  to  meet  EF  which  has  been  drawn  through  D j | 
AB.  With  K as  a center  and  KA  as  a radius,  and 
E as  a center  and  EG  as  a radius,  describe  arcs 
meeting  at  G.  Prove  that  these  arcs  have  a com- 
mon tangent  at  G,  and  therefore  form  a compound 
curve.  (See  Hanstein’s  Constructive  Drawing.) 


35.  Construct  a Persian  arch  in  which  the  arc  DG  = arc  NG. 

[Sug.  Bisect  line  AD  instead  of  trisecting  it.] 

36.  Construct  Persian  arches  in  which  the  chords  NG  and  DG  have 


PRACTICAL  APPLICATIONS 


519 


various  ratios,  and  decide  which  of  these  arches  you  think  is  the  most 
beautiful. 

37.  Construct  segmental  arches  of  various  shapes  and  decide  which 
of  these  you  think  is  the  most  beautiful. 

38.  Construct  a trefoil,  given  the  radius  of  one  of  its  circles. 

[Sug.  Construct  an  equilateral  triangle,  one  of  whose 

sides  equals  twice  the  given  radius.  Take  each  vertex 
of  the  triangle  as  a center,  and  half  of  one  of  the  sides 
as  a radius  and  describe  arcs.] 

39.  Inscribe  a trefoil  in  a given  equilateral  triangle. 

[Sug.  Bisect  the  angles  of  the  triangle,  etc.] 

40.  Discover  the  method  of  construction  of  each  of  the  figures  or 
diagrams  on  the  next  page,  and  then  reproduce  each  of  them  in  a 
drawing. 

Squared  paper  may  be  used  to  advantage  as  an  aid  in  making  some 
of  these  constructions.  See  Fig.  10,  p.  520. 

41.  Construct  a diagram  similar  to  Fig.  10,  but  making  a rectangle 
instead  of  a square  the  basis  of  the  drawing. 

42.  Also  one  making  the  rhomboid  the  basis. 

43.  Construct  a trefoil  and  develop  into  an  ornamental  design  by 
placing  small  circles  and  arcs  within  its  parts  and  larger  circles  outside. 

44.  By  use  of  squared  paper  invent  designs  similar  to  Fig.  10  on 
p.  520,  but  formed  by  two  intertwining  lines. 

46.  Collect  a number  of  pictures  of  ornamental  designs,  tracery, 
scroll  work,  etc.,  such  as  are  used  in  architecture,  wall  paper,  and 
similar  patterns,  whose  construction  depends  on  the  principles  of  Book 
II.  Show  how  these  designs  are  constructed,  and  reproduce  them  in 

EXERCISES.  GROUP  90 
(Book  III) 

1.  To  find  the  distance  between  two  accessible 
objects  A and  B separated  by  a barrier,  take  a 
convenient  point  C,  measure  AC  and  BC.  Produce 
BC  to  D and  AC  to  F so  that  DC  — some  frac- 
tion, as  o,  of  CB,  and  FC  = the  same  fraction  of 
AC,  and  measure  FD.  How,  then,  is  AB  com- 
puted? Prove  this. 


drawings. 


520 


GEOMETRY 


PRACTICAL  APPLICATIONS 


521 


Let  BC  = 300  yd.,  DC  = 60  yd.,  AC  = 240  yd.,  FC  = 48  yd., 
FD  = 52  yd.,  find  AB. 

Why  is  this  method  of  finding  AB  often  more  convenient  than  that 
of  Ex.  7,  p.  511. 

2.  What  similar  method  does  the  adjoining  dia- 
gram suggest  for  finding  AB? 

3.  In  the  diagram  of  Ex.  2,  in  case  a river  ran 
between  A and  B,  and  also  between  B and  F,  what 
measurements  would  be  necessary  in  order  to  determine  the  length 
of  AB?  Of  BF? 

4.  Work  again  Ex.  2,  p.  310. 

5.  In  case  the  sun  is  not  shining  and  shadows  cannot  be  used,  the 

height  of  an  object  like  a 
tree  or  steeple  can  often  be 
determined  by  a method  in- 
dicated in  the  drawing. 
What  distance  must  be  meas- 
ured and  why  to  determine 
the  height  of  the  tree? 

6.  Show  how  to  find  the 
height  of  a tree  by  placing 
a mirror  in  a horizontal  posi- 
tion on  the  ground  and  stand- 
ing so  as  to  see  the  reflection 
of  the  top  of  the  tree  in  the  mirror.  If  the  observer’s  eye  is  5J  ft. 
from  the  ground,  the  observer  stands  6 ft.  from  the  mirror,  and 
the  mirror  is  120  ft.  from  the  tree, 
how  high  is  the  tree? 

7.  Foresters  often  determine  the 
height  of  a tree  by  an  instrument 
called  “Faustman’s  Height  Meas- 
urer.” The  principle  on  which  this 
instrument  is  constructed  is  shown 
in  the  diagram. 

If  the  distance  from  A to  the  foot 
of  the  tree  is  150  ft.,  BC  = 6 inches, 
and  CF  = 4|  inches,  find  the  height 
of  the  tree. 

8.  The  distance  from  A to  B in  Ex.  7,  p.  511,  might  have  been  deter- 
mined by  a graphical  method  as  follows:  Measure  AC,  CB,  and  angle 


522 


GEOMETRY 


ACB.  On  paper  make  a drawing  of  the  triangle  ACB  to  a convenient 
scale.  On  this  drawing  measure  the  line  which  represents  AB  and 
hence  determine  the  length  of  AB. 

By  use  of  this  method  we  are  saved  the  labor  of  marking  out  and 
measuring  the  lines  CD,  CF,  and  DF. 

Apply  this  method  to  the  measurement  of  two  objects  in  your  neigh- 
borhood which  are  separated  by  an  impassable  barrier. 

In  like  manner  show  how  the  distance  from  a given  point  to  another 
inaccessible  place  may  be  determined. 

9.  Also  the  distance  between  two  places  both  of  which  are  inacces- 
sible. 


10.  What  is  the  measuring  instrument  called  the  diagonal  scale? 
How  is  the  principle  of  similar  triangles  used  in  it?  Show  how  the 
diagonal  scale  aids  in  the  accurate  measurement  of  lines  and  hence  in 
the  accurate  determination  of  a distance  such  as  AB  in  Ex.  7,  p.  511. 


q 11.  In  the  triangle  OAB, 
/ A is  a right  angle,  and  OA 
s'  is  1.  By  a method  which 
/ is  beyond  the  scope  of  this 

y'  book,  the  length  of  AB  is 

/fa  computed  and  found  to  be 

P 250  it.  R .839  +.  Using  this  fact  find 
RQ  in  the  second  triangle. 


12.  By  use  of  the  accompanying  table,  find  RQ  if  angle  P is  10°.  20°. 
70°.  80°. 


Tables  giving  the  other  sides  of  all  possible  right  triangles  when  one 
side  is  unity  have  been  computed  and  when  used  as  in  Exs.  11  and  12, 
form  the  basis  of  the  subject  of  Trigo- 
nometry. By  use  of  this  science,  after 
measuring  the  length  of  a single  fine  a 
few  miles  long  on  the  earth’s  surface,  we 
can  determine  the  distances  and  relative 
positions  of  other  places  thousands  of 
miles  away,  without  measux-ing  any  in- 
tervening lines.  By  use  of  these  results 
as  a basis,  the  distance  of  the  moon  is 
determined  as  approximately  240,000 
miles;  of  the  sun  as  92,800,000  miles; 
and  of  the  nearest  fixed  star  as  20,000,- 
000,000,000  miles. 


Angle  O 

AB 

OB 

10° 

.176 

1.015 

20° 

.364 

1.064 

30° 

.577 

1.155 

40° 

.S39 

1.305 

50° 

1.192 

1.556 

60° 

1.732 

2.000 

70° 

2.747 

2.924 

o 

O 

00 

5.671 

5.759 

PRACTICAL  APPLICATIONS  523 


A knowledge  of  these  distances  has  led  to  important  improvements 
in  methods  of  navigation  and  have  thus  facilitated  travel  and  com- 
merce and  increased  their  benefits  for 
us  all. 

13.  On  the  diagram  given  AB  is  6,000  ft., 
and  the  angles  as  indicated,  compute  the 
length  of  CD,  by  use  of  the  table  given  in 

Ex.  12. 

14.  To  the  diagram  in  Ex.  13  annex 
another  triangle  CFD  giving  it  angles  which 
are  multiples  of  10°,  and  compute  the 
length  of  CF. 

16.  A pantograph  is  an  instrument  for 
drawing  a plane  figure  similar  to  a given 
plane  figure.  It  is  used  for  enlarging  or 
reducing  maps  and  drawings.  It  consists  of  four  bars,  parallel  in 
pairs  and  jointed  at  c,  b,  C,  and  E,  as  shown  in  the  diagram.  cbEC 
is  a parallelogram.  The  rods  may  be  joined  so  that  any  required 


ratio  = 


A turns  upon  a 


Ac  _ CE 
AC  CB' 

fixed  pivot,  and  pencils  are  carried 
at  b and  F. 

Show  that  A,  b,  and  B are  always 

Ab  , , , . .Ac 

in  a straight  line,  and  that  — — always  equals  the  given  ratio 


AB 


AC 


[Sttg.  Draw  a line  from  A to  b,  and  a line  from  A to  B.  Prove  the 
triangle  A cb  and  ACB  similar  (Art  193);  hence  show  that  Ab  and  AB 
coincide,  etc.] 

16.  Using  the  fact  that  a triangle  whose  sides  are  3,  4,  and  5 units  of 
length  is  a right  triangle,  show  how,  by  stretching  a 100-ft.  tape,  to 
construct  a right  angle  as  accurately  as  possible.  (Among  the  ancient 
Egyptians  a class  of  workmen  existed  called  rope  stretchers,  whose 
business  was  to  construct  right  angles  in  this  general  way.) 

17.  The  strongest  beam  which  can  be  cut  from  a 
given  round  log  is  found  as  follows:  Take  AB,  a 
diameter  of  the  log,  and  trisect  it  at  C and  D.  Draw 
CE  and  DF  JL  AB  and  meeting  the  circumference  at 
E and  F respectively.  Draw  AF,  FB,  BE,  and  AE. 

Prove  AFBE  a rectangle;  also  FB:AF  = 1:  y/2, 
or  approximately,  as  5:7. 


524 


GEOMETRY 


[Sug.  FB  is  a mean  proportional  between  DB  and  AB.  Therefoie 

FB2  — 1 AB 2 (Art.  343).] 

In  like  manner  FA2  = § AB2,  etc.] 

18.  A sphere  S weighing  100  lb.  rests  on  the  inclined  plane  AB.  AC 
contains  8 units  of  length,  BC  6 units.  A 
force  which  prevents  S from  rolling  down 
the  plane  would  be  equivalent  to  a lifting 
force  of  how  many  pounds  exerted  on  S and 
parallel  with  A B? 

[Sug.  Resolve  the  weight  of  S (represented 
by  SP)  into  two  forces,  one  perpendicular  to 
AB  and  the  other  parallel  to  AB.  Prove  the 
triangles  ACB  and  SPR  similar,  and  obtain 
AB:  BC  = SP:  SR,  or  10:  6 = 100  lb.:  SR.] 

The  principle  involved  in  this  example  is  of  great  practical  impor- 
tance. Thus  in  many  machines  useful  results  are  often  obtained  by 
representing  a force  by  the  diagonal  of  a rectangle  (or  parallelogram) 
and  separating  this  force  into  two  component  forces  represented  by 
the  sides  of  the  rectangle  (or  parallelogram),  only  one  of  these  com- 
ponents being  effective.  This  principle  makes  possible  the  action  of  the 
propeller  of  an  aeroplane  or  steamboat,  of  the  best  water  wheels  and 
wind  mills,  and  indeed  of  all  turbine  wheels.  It  also  determines  the 
lifting  power  of  the  planes  of  an  aeroplane. 

19.  A wagon  weighing  1800  lb.  stands  on  the  side  of  a hill  which  has 
a rise  of  18  ft.  for  every  100  ft.  taken  horizontally.  Hence  what  force 
must  a horse  exert  to  keep  such  a wagon  from  running  down  hill,  fric- 
tion being  neglected? 

20.  Make  up  and  work  a similar  example  for  yourself. 

21.  Show  how  the  diameter  of  the  earth  may  be  determined  by  the 
following  method:  Drive  three  stakes  in  a level  piece  of  ground  (or  in  a 
shallow  piece  of  water)  in  line,  each  two  successive  stakes  being  a mile 
apart,  and  let  each  stake  project  the  same  distance  above  the  ground 
(or  water).  By  use  of  a leveling  instrument,  determine  the  amount 
by  which  the  middle  stake  projects  above  a horizontal  line  connecting 
the  tops  of  the  end  stakes.  This  distance  will  be  found  to  be  8 in. 

[Sug.  Use  Art.  343.  An  arc  a mile  long  on  the  earth’s  surface  may 
be  taken  as  equal  to  its  chord.  Then  from  the  diagram  of  Art.  343  we 
obtain  the  following  proportion, 

the  diameter  of  the  earth.  1 mi.  = 1 mi.:  8 in.] 

22.  Also  show  that  the  distance  that  the  middle  stake  projects  above 


PRACTICAL  APPLICATIONS 


525 


the  horizontal  line  connecting  the  top  of  the  two  end  stakes  varies  as 
the  square  of  the  distance  between  the  end  stakes.  Thus  if  the  two  end 
stakes  were  placed  three  times  as  far  apart  as  in  Ex.  21  (that  is,  6 miles 
apart  instead  of  2 miles)  the  bulge  of  the  earth  between  them  would 
be  32  or  9 times  what  it  was  originally. 

Hence  determine  the  projection  of  the  middle  stake  (or  bulge  of  the 
earth)  when  the  end  stakes  are  4 mi.  apart.  Also  8 mi.  16  mi.  32  mi. 

23.  At  the  seashore  an  observer  whose  eye  was  10  ft  above  sea  level 
observed  a distant  steamboat  whose  hull  was  hidden  for  a height  of 
12  ft.  above  water  level  by  the  bulge  of  the  earth.  About  how  far  off 
was  the  steamboat? 

24.  A seaman  in  a lookout  42  feet  above  water  level  with  a glass 
could  barely  see  the  topsail  of  a distant  ship,  and  estimated  this  top- 
sail to  be  45  ft.  above  sea  level.  Estimate  the  distance  of  the  observed 
vessel  from  the  seaman. 

25.  Work  again  Exs.  7-8,  p.  310,  Group  56. 

26.  The  moon’s  distance  from  the  earth’s  center  approximately 
equals  60  times  the  earth’s  radius.  A body  falling  at  the  earth’s  sur- 
face goes  193  in.  in  1 sec.  Hence,  if  the  law  of 
gravitation  is  true,  the  distance  the  moon  falls 

193  in. 

in  1 sec.  toward  the  earth  will  be In  the 

602 

diagram,  let  0 be  the  earth’s  center,  ABE  the 
moon’s  orbit,  and  CB  or  AD  the  distance  the 
moon  falls  toward  the  earth  in  1 sec.  Taking 
the  month  as  27  da.  7 hr.  43  min.  11  sec., 

show  that  AD  = approximately.  This  is  the  calculation  used 

602 

by  Sir  Isaac  Newton  in  testing  the  truth  of  the  law  of  gravitation. 

27.  Any  rectangular  object,  as  a book,  door,  or  photograph,  is  con- 
sidered to  be  of  the  most  artistic  shape  when  its  length  and  breadth 
have  the  same  ratio  as  the  segments  of  a line  divided  in  mean  and 
extreme  ratio  (Art.  370).  In  accordance  with  this  rule,  if  a window  is 
6 feet  high,  how  wide  should  it  be? 

The  division  of  a line  in  extreme  and  mean  ratio  has  been  termed  the 
Golden  Section.  By  many  the  Golden  Section  is  regarded  as  a fundamen- 
tal principle  of  esthetics,  having  applications  in  determining  the  propor- 
tions of  the  ideal  human  form,  in  explanation  of  musical  harmonies,  etc. 

28.  Make  up  and  work  an  example  similar  to  Ex.  27. 


526 


GEOMETRY 


EXERCISES.  GROUP  91 


(Book  IV) 

1.  The  supporting  power  of  a wooden  beam  (of  rectangular  cross 
section  and  of  given  length)  varies  as  the  area  of  the 
cross  section  multiplied  by  the  height  of  the  beam. 
If  the  cross  section  of  a given  beam  is  4 " X 8 ",  com- 
pare the  supporting  power  of  the  beam  when  it  rests 

° on  the  narrow  edge  (4")  with  its  supporting  power 
when  it  rests  on  its  wide  edge  (8"). 

2.  Two  beams  of  the  same  length  and  material  have  cross  sections 
which  are  2"  X 4"  and  3"  X 8"  respectively.  Find  the  ratio  of  the 
supporting  power  of  the  two  beams. 

3.  In  Ex.  17,  p.  523,  compare  the  supporting  power  of  a beam  cut 
from  a log  in  the  manner  indicated,  with  that  of  a square  beam  cut 
from  the  same  log. 

4.  Also  with  that  of  a beam  whose  width  equals  I of  the  diameter. 

6.  When  an  irregular  area  like  ABCD  is  ^ /_1^  j5 

calculated  by  means  of  equidistant  offsets,  the 
following  rule  is  used:  To  the  half  sum  of  the 
initial  and  final  offsets  add  the  sum  of  all  the 
intermediate  offsets,  and  multiply  the  sum  by 
between  the  offsets.  Prove  this  rule. 

6.  Show  how  the  rule  of  the  preceding  prob- 
lem could  be  used  to  calculate  an  area  whose 
entire  boundary  is  an  irregular  curved  line. 

7.  Surveyors  often  determine  jy 

the  area  of  a piece  of  land,  as  of  ABCD,  by  taking 
an  auxiliary  line  as  NS,  measuring  the  perpendicular 
distances  from  A,  B,  C,  D,  E to  NS,  and  the  inter- 
cepts on  NS  between  these  perpendiculars,  and 
combining  the  areas  of  the  various  trapezoids  (or  k 
triangles)  formed.  Supply  probable  numbers  for  the 
lengths  of  lines  on  the  diagram,  and  compute  the 
area  of  ABODE.  ° 

d a,  a „ o,  t Frequently  (as  when  the  center  of 

the  curve  cannot  be  seen  from  the  curve) 
a railroad  curve  is  laid  out  by  construc- 
o>\C  ting  a series  of  equidistant  offsets  per- 
pendicular to  the  tangent  of  the  curve. 


common  distance 


PRACTICAL  APPLICATIONS 


527 


Thus,  if  ABT  is  a straight  track  and  BC  is  a curve  to  be  laid  out  tan- 
gent to  AB  at  B,  mark  off  Bai,  aia2,  0,20,3,  all  = d,  and  construct  the  _L 
offsets  Chbi,  a2&2,  0363  by  using  the  formula  anbn  = r — v r2  — n-d 2 
where  r is  the  radius  of  the  curve.  Prove  that  this  formula  is  correct. 


9.  The  diagram  represents  two  straight  parallel  railroad  tracks 
connected  by  a compound  curve  (in  this  v. 

case  called  a cross-over  track)  composed  of  pj- - 

two  tracks  with  common  tangents  at  D and 
E with  centers  0 2 and  Oi,  and  having  equal 
radii.  Taking  the  magnitudes  as  indicated 
on  the  figure,  show  that  KPOJD  is  a rect- 
angle (use  Arts.  122,  160). 

10.  Find  a formula  for  r in  terms  of  l, 
d and  w.  (Use  the  right  triangle  0\  P02 
and  Art.  400.)  Also  for  l in  terms  of 
r,  d,  and  w. 

11.  If  w = 4'  82",  d = 10  ft.,  r = 150  ft., 
find  l. 


12.  If  a steamboat  is  traveling  at  the  rate  of  12  miles  an  hour,  and  a 
boy  walks  across  her  deck  at  right  angles  to  her  line  of  motion  at  the 
rate  of  3 miles  an  hour,  draw  a diagram  to  show  the  direction  of  his 
resultant  motion.  From  this  determine  the  speed  at  which  he  is  going. 

13.  Make  and  work  a similar  example  for  yourself  concerning  a mail 
bag  thrown  from  a train. 

14.  Also  concerning  a breeze  blowing  into  a window  of  a moving 
trolley  car. 

15.  Two  forces,  one  of  300  lb.,  the  other  of  400  lb.,  act  at  right  angles 
on  the  same  body.  Find  their  resultant. 

16.  Make  up  and  work  a similar  example  for  yourself. 

17.  A river  is  flowing  at  a rate  of  4.25  mi.  per  hour,  and  a man  is 
rowing  at  right  angles  with  the  current  at  a rate  of  3.75  mi.  per  hour. 
What  is  the  resultant  velocity  of  the  man? 

18.  If  a star  has  a velocity  of  15  mi.  a second  toward  the  earth  and  a 
velocity  of  20  mi.  a second  at  right  angles  with  a line  drawn  from  the 
star  to  the  earth,  find  the  velocity  of  the  star  in  its  own  path. 

19.  A body  is  moving  in  the  straight  line  AD  past  the  point  O (p.  528). 
The  distance  passed  over  in  a unit  of  time  = AB  = BC  = CD.  When 
the  body  reaches  B it  is  acted  on  by  a force  which  impels  it  toward  O 


528 


GEOMETRY 


a distance  BF  in  a unit  of  time.  Prove  that  the  area  swept  over  in  the 
unit  of  time  by  a line  drawn  from  the  body  to  0 will  be  unchanged 
by  the  action  of  the  new  force.  (That  is,  on  the 
diagram  prove  that  AOCB  =0=  AOPB.) 

This  principle  accounts  for  the  fact  that  the  earth 
(or  another  planet  or  a comet)  moves  faster  in  its 
orbit  the  nearer  it  is  to  the  sun. 

20.  By  tracing  the  outline  of  a map  on  squared 
paper,  show  how  to  find  the  area  of  an  irregular 
figure  as  of  some  country  or  part  of  a country.  By 
use  of  this  method  find  the  area  of  some  part  of 
I \j  the  state  in  which  you  five. 

EXERCISES.  GROUP  92 

(Book  V) 

1.  A cooper  in  fitting  a head  to  a barrel  takes  a pair  of  compasses 
and  then  adjusts  them  till,  when  applied  six  times  in  succession  in  the 
chine,  they  will  exactly  complete  the  circumference.  He  then  takes 
the  distance  between  the  points  of  the  compasses  as  the  radius  of  the 
head  of  the  barrel.  Why  is  this? 

2.  Draw  a square  and  convert  it  into  a regular  octagon  by  cutting 
off  the  corners. 

[Sug.  Draw  the  diagonals  of  the  square,  bisect  their  angles  of  inter- 
section, etc.] 

3.  In  heating  a house  by  a hot-air  furnace  the  area  of  the  cross 
section  of  the  cold-air  box  should  equal  the  sum  of  the  areas  of  the  pipes 
conducting  hot  air  from  the  furnace.  If  a given  furnace  has  three  hot- 
air pipes,  each  6"  in  diameter,  and  one  pipe  8"  in  diameter,  and  the 
width  of  the  cold-air  box  is  15  ",  how  deep  should  the  box  be? 

4.  What  is  the  most  convenient  way  of  determining  the  diameter  of 
a hot-air  pipe  if  you  have  no  callipers  and  the  ends  of  the  pipe  are  not 
accessible? 

5.  A half-mile  running  track  is  to  have  equal  semicircular  ends  and 
parallel  straight  sides.  The  extreme  length  of  the  rectangle  together 
with  the  semicircular  ends  is  to  be  1000  ft.  Find  the  width  of  the 
rectangle. 

[Sug.  Denote  the  length  of  the  radius  of  the  semicircular  ends  by 
x and  that  of  one  of  the  parallel  side  straight  tracks  by  y,  and  obtaiif 
a pair  of  simultaneous  equations.] 


PRACTICAL  APPLICATIONS 


529 


6.  A belt  runs  over  two  wheels  one  of  which  has  a diameter  of  3 ft. 
and  the  other  of  6 in.  If  the  first  wheel  is  making  120  revolutions  per 
minute,  how  many  is  the  second  wheel  making.  How  many  revolu- 
tions must  the  first  wheel  make  in  order  that  the  second  may  make 
300  revolutions  per  minute? 

7.  If  in  laying  a track  a rail  10  ft.  long  is  bent  through  an  angle  of 
5°  10',  wrhat  is  the  radius  of  the  curve? 

Assuming  (what  is  not  strictly  true,  owing  to  friction  against  the 
sides  of  the  pipe,  etc.)  that  the  rate  of  flow  through  a cylindrical  pipe  is 
proportional  to  its  area  of  cross  section: 

8.  Work  again  Ex.  12,  p.  277. 

9.  If  a l|-in.  pipe  is  replaced  by  a 1-in.  pipe,  how  much  is  the  flow 
of  water  increased? 

10.  A 3-in.  pipe  is  to  be  replaced  by  one  which  will  deliver  twice  as 
much  water  per  minute.  Find,  to  the  nearest  quarter  of  an  inch,  the 
diameter  of  the  new  pipe. 

11.  A 2-in.  steam  pipe  conveying  steam  from  the  boiler  to  the  radia- 
tors in  a school  building  is  found  to  supply  only  two  thirds  the  needed 
amount  of  steam.  What  is  the  smallest  even  size  of  pipe  that  will 
convey  the  needed  amount? 

12.  A city  of  40,000  people  is  barely  supplied  with  water  by  12-in. 
mains  from  the  reservoirs.  If  these  mains  are  torn  out,  and  18-in. 
mains  substituted,  what  future  population  of  the  city  is  allowed  for? 

13.  A steel  bar  1 in.  in  diameter  will  hold  up  50,000  lb.  What  load 
would  a bar  f in.  in  diameter  carry?  What  would  be  the  diameter  of  a 
round  (cylindrical)  bar  to  carry  150,000  lb.? 

14.  If  the  center  of  symmetry  of  a flat,  homogeneous  object  is  the 
center  of  mass,  find  the  center  of  mass  of  a square;  of  a rectangle; 
regular  hexagon;  circle. 

15.  It  is  evident  that  if  the  medium  of  a triangle  (BM)  be  placed  on 
a knife  edge  the  triangle  will  balance  (for  if  PP' 
be  ||  AC,  the  pull  on  P is  balanced  by  the  pull  on 
P').  Hence  find  the  center  of  mass  for  any  triangle. 

For  a regular  pentagon. 

It  is  useful  to  be  able  to  determine  the  center  of 
mass  of  an  object  by  geometry  or  by  any  other 
means,  since  a knowledge  of  the  center  of  mass  of  a body  often 
enables  us  to  treat  the  body  in  a simple  way,  for  example,  as  if  the 
body  were  concentrated  at  a single  point. 


530 


GEOMETRY 


16.  If  a box  has  a triangular  end,  subject  to  the  same  pressure  at  all 
points,  at  what  single  point  on  the  end  must  a supporting  pressure 
be  applied? 


17.  The  cross  section  of  a cylinder  is  a circle.  The  weight-supporting 
strength  of  a horizontal  cylindrical  beam  of  given  material  and  length 
varies  as  the  area  of  the  cross  section  times  its  radius. 

Compare  the  weight-supporting  power  of  two  solid  horizontal  iron 
cylindrical  beams  of  the  same  length  and  quality  of  iron,  the  radii 
being  3 in.  and  6 in.  respectively.  (Point  out  and  use  the  short  way 
of  getting  the  desired  result.) 


18.  The  cross  section  of  a hollow  cylinder  (i.e.  of  a tube)  is  a circular 
ring.  Denote  the  outside  radius  of  the  tube  by  R and  the  inside  radius 
by  r.  Then  it  may  be  shown  that  the  weight-supporting  power  of  a 
hollow  cylindrical  tube  of  given  length  and  material  varies  as  the  area 

R?  + r2 


of  the  cross  section  (i.e.  of  the  ring)  times - 


R 


If  R = 4 in.  and  r = 3 in.,  compare  the  weight-supporting  power 
of  the  tube,  with  that  of  a solid  cylindrical  beam  of  the  same  length 
and  same  cross  sectional  area. 

19.  Make  up  and  work  a similar  example. 

In  general  a cylindrical  tube  is  stronger  than  a solid  cylindrical  beam 
of  the  same  length  and  containing  the  same  amount  of  material. 

Hence  in  a framework,  as  in  that  of  an  airship  where  the  maximum 
strength  must  be  obtained  from  a given  amount  of  material,  the  metallic 
rods  and  posts  are  all  tubular.  In  like  manner,  bamboo  rods,  since 
they  are  hollow,  are  used  in  an  aeroplane  instead  of  solid  wooden  rods 
wherever  possible.  For  the  same  reason,  the  bones  of  flying  birds,  and 
many  bones  in  men  and  animals  are  hollow  and  not  solid. 

20.  To  construct  a square  which  shall  be  ap- 
proximately equivalent  to  a given  circle  0,  divide 
the  radius  OA  into  four  equal  parts  and  produce 
each  end  of  two  perpendicular  diameters  a dis- 
tance equal  to  one  fourth  of  the  radius,  and 
connect  the  extremities  of  the  lines  thus  formed. 
Show  that  taking  the  square  thus  formed  as 
equivalent  to  the  circle  is  the  same  as  taking 
v = 3y.  Also  find  the  per  cent,  of  error  in  taking  the  square  as 
equivalent  to  the  circle. 

21.  A short  way  to  construct  a regular  inscribed  pentagon  and  also 
a five-pointed  star  (or  pentagram)  is  as  follows:  Draw  a circle  0 and 


PRACTICAL  APPLICATIONS 


531 


two  diameters  AB  and  CD  at  right  angles.  Bisect  the  radius  OB  at  F, 
and  with  F as  a center  and  FC  as  a radius  describe  an  arc  cutting  AO 
at  H.  Then  CH  is  the  length  of  a side  of  c 

the  regular  inscribed  pentagon,  by  joining 
whose  alternate  vertices  the  pentagram 
may  be  formed. 

As  a proof  of  this,  by  use  of  right  tri- 
angles, prove 


CH  = R 


V10-2V5 


(See  Ex.  17,  p.  301.) 


r\ 

j 

D 


22.  Carpenters  some- 
times use  the  following 

method  of  approximately  determining  the  length 
of  a circumference  whose  radius  is  known:  Let 
0 be  the  given  circle.  Draw  OA  and  OB,  radii 
at  right  angles.  Draw  AB  and  the  radius  OEA.AB 
at  D.  Measure  DE.  Then  take  circumference 
= 6 AO  + DE. 

Find  the  per  cent,  of  error  in  this  method,  taking  t =>  3.14159-. 


23.  The  following  designs  are  important  in  architecture  or  in  orna- 
mental work  (thus  the  first  is  a detail  in  a stained  glass  window  in  an 
early  French  cathedral;  the  second  is  the  plan  of  the  base  of  a col- 
umn in  an  English  cathedral).  Discover  a method  of  constructing  each 
of  the  following  designs,  and  reproduce  each  of  them  in  a drawing: 


532 


GEOMETRY 


24.  Construct  a square.  Take  each  vertex  of  the  square  as  a center 
and  one  half  a side  of  the  square  as  a radius  and  describe  arcs  which 
meet.  Erase  the  square  and  you  have  a quatrefoil.  By  drawing 
other  circles  and  arcs  of  circles,  elaborate  the  quatrefoil  into  an  orna- 
mental design  (see  design  11,  p.  520). 

25.  In  like  manner  construct  a cinquefoil  by  use  of  a regular  penta- 
gon, and  develop  it  into  an  ornamental  design. 

26.  Treat  a regular  hexagon  in  the  same  way. 

27.  A pavement  or  mosaic  may  be  formed  out  of  regular  polygons  in 
the  following  ways.  Show  how  to  make  each  diagram  in  the  simplest 
way. 


APPLICATIONS  OF  SOLID  GEOMETRY  TO  MECHANICS 
AND  ENGINEERING 


EXERCISES.  GROUP  93 


(Books  VI  and  VII) 

1.  A carpenter  tests  the  flatness  of  a surface  by  applying  a straight 
fcdge  to  the  sin-face  in  various  directions.  How  does  a plasterer  test 
the  flatness  of  a wall  surface?  What  geometric  principle  is  used  by 
these  mechanics? 

2.  Explain  why  an  object  with  three  legs,  as  a stool  or  tripod, 
always  rests  firmly  on  the  floor  while  an  object  with  four  legs,  as  a table, 
does  not  always  rest  so.  Why  do  we  ever  use  four-legged  pieces  of 

30'  furniture? 

3.  How  can  a carpenter 
get  a comer  post  of  a house 
in  a vertical  position  by  use  of 
a carpenter’s  square?  What 
geometrical  principle  does  he 
use? 

4.  The  diagram  is  the  plan 
of  a hip  roof.  The  slope  of 
each  face  of  the  roof  is  30°. 
Find  the  length  of  a hip  rafter 
as  AB. 

[Sug.  Draw  a triangle  DBC  representing  a section  of  the  roof  at 

20 

DBC  on  the  plan.  Hence  it  may  be  shown  that  BC  = — - x/3.  In 

* O 

like  manner  by  taking  a section  through 
BF,  it  is  found  that  AC  = 10.  Hence  in 
the  triangle  ABC,  AB  may  be  found.] 

5.  Find  the  area  of  the  entire  roof 

represented  in  Ex.  4.  p 

6.  Make  drawings  showing  at  what 
angle  the  two  ends  of  a rafter  like  BC  in  Ex.  4 must  be  cut. 

533 


534 


GEOMETRY 


7.  Make  drawings  showing  at  what  angle  a jack  rafter  like  12  in 
Ex.  4 must  be  cut. 

[Sug.  To  determine  how  the  end  1 of  the  jack  rafter  must  be  cut 
use  the  principle  that  two  intersecting  straight  lines  determine  a plane 
(Art.  501).  The  cutting  plane  at  1 must  make  an  angle  at  the  side 
of  the  jack  rafter  equal  to  angle  CBH,  and  on  the  top  of  the  jack  rafter 
equal  to  angle  ABC.] 

8.  What  is  a gambrel  roof?  Make  up  a set  of  examples  concerning 
a gambrel  roof  similar  to  Exs.  4-7. 

9.  By  use  of  Art.  645,  show  that  a page  of  this  book  held  at  twice 
the  distance  of  another  page  from  the  same  lamp  receives  one  fourth 
the  light  the  first  page  receives. 

10.  The  supporting  power  of  a wooden  beam  varies  directly  as  the 
area  of  the  cross  section  times  the  height  of  the  beam  and  inversely  as 
the  length  of  the  beam.  Compare  the  supporting  power  of  a beam 
12  ft.  long,  3 in.  wide,  and  6 in.  high  with  that  of  a beam  18  ft.  long,  4 in, 
wide,  and  10  in.  high.  Also  compare  the  volumes  of  the  two  beams. 


EXERCISES.  GROUP  94 
(Books  VIII  and  IX) 

1.  A hollow  cylinder  whose  inside  diameter  is  6 in.  is  partly  filled 
with  water.  An  irregularly  shaped  piece  of  ore  when  placed  in  the 
water  causes  the  top  surface  of  the  water  to  rise  3.4  in.  in  the  cylinder. 
Find  the  volume  of  the  ore. 

2.  What  is  a tubular  boiler?  What  is  the  advantage  in  using  a 
tubular  boiler  as  compared  with  a plain  cylindrical  boiler?  If  a tubular 
boiler  is  18  ft.  long  and  contains  32  tubes  each  3 in.  in  diameter,  how 
much  more  heating  surface  has  it  than  a plain  cylindrical  boiler  of  the 
same  length  and  36  in.  in  diameter?  (Indicate  both  the  long  method  and 
the  short  method  of  making  this  computation  and  use  the  short  method.) 

3.  If  a bridge  is  to  have  its  linear  dimensions  1000  times  as 
great  as  those  of  a given  model,  the  bridge  will  be  how  many  times 
as  heavy  as  the  model? 

Why,  then,  may  a bridge  be  planned  so  that  in  the  model  it  will  sup- 
port relatively  heavy  weights,  yet  when  constructed  according  to  the 
model,  falls  to  pieces  of  its  owm  weight? 

Show  that  this  principle  applies  to  other  constructions,  such  as 
buildings,  machines,  etc.,  as  well  as  to  bridges. 


PRACTICAL  APPLICATIONS 


535 


4.  Work  again  Exs.  23-25,  27,  p.  473. 

6.  Make  and  work  for  yourself  an  example  similar  to  Ex.  24,  p.  473. 

6.  Sound  spreads  from  a center  in  the  form  of  the  surface  of  an 
expanding  sphere.  At  the  distance  of  10  yd.  from  the  source,  how  will 
the  surface  of  this  sphere  compare  with  its  surface  as  it  was  at  1 yd.? 
How,  then,  does  the  intensity  of  sound  at  10  yd.  from  the  source  com- 
pare with  its  intensity  at  a distance  of  1 yd.? 

Does  this  law  apply  to  all  forces  which  radiate  or  act  from  a center 
as  to  light,  heat,  magnetism,  and  gravitation?  Why  is  it  called  the 
law  of  inverse  squares? 


7.  If  a body  be  placed  within  a spherical  shell,  the  attractive  forces 
exerted  upon  the  body  by  different  parts  of  the  shell  will  balance  or 
cancel  each  other.  Hence  a body  inside  the  earth,  as  at  the  foot  of  a 
mine,  is  attracted  effectively  only  by  the  sphere  of  matter  whose  radius 
is  the  distance  from  the  center  of  the  earth  to  the  given  body.  Hence, 
prove  that  the  weight  of  a body  below  the  surface  of  the  earth  varies 
as  the  distance  of  the  body  from  the  center  of  the  earth. 

[Sug.  If  W denote  the  weight  of  the  body  at  the  surface,  and  w its 
weight  when  below,  R the  radius  of  the  earth,  and  r the  distance  of  the 

body  from  the  center  when  below 
the  surface,  show  that 

m R?  r*  „ 

w-w 

8.  The  light  of  the  sun  falling 
on  a smaller  sphere,  as  on  the  earth  or  the  moon,  causes  that  body  to 
cast  a conical  shadow.  Denoting  the  radius  of  the  sun  by  R,  the 
radius  of  the  smaller  sphere  by  r,  the  distance  between  the  two  sphere# 

d r 

by  d,  and  the  length  of  the  shadow  by  l,  show  that  l = — . 

R — r 

Find  l when  d - 92,800,000  mi.,  R — 433,000  mi.,  r = 4000  mi. 


9.  If  in  a lunar  eclipse  the  moon’s  center  should  pass  through  the 
axis  of  the  conical  shadow,  and  the  moon  is  traveling  at  the  rate  of 
2100  mi.  an  hour,  how  long  would  the  total  eclipse  of  the  moon  last? 

How  long  if  the  moon’s  center  passed  through  the  earth’s  conical 
shadow  at  a distance  of  1000  mi.  from  the  axis  of  the  cone? 


536 


GEOMETRY 


10.  If  the  moon’s  diameter  is  2160  mi.,  find  the  length  of  the  moon’s 
shadow  as  caused  by  sunlight. 

11.  If  the  distance  of  the  moon  from  the  earth’s  center  varies  from 
221,600  mi.  to  252,970,  show  how  this  explains  why  some  eclipses  of  the 
sun  are  total  and  others  annular. 

Why,  also,  at  a given  point  on  the  earth’s  surface  is  an  eclipse  of  the 
sun  a so  much  rarer  sight  than  an  eclipse  of  the  moon?  Why,  also,  is  its 
duration  so  much  briefer? 


FORMULAS  OF  PLANE  GEOMETRY 

SYMBOLS 


a,  b,  c=  sides  of  triangle  ABC. 
s=i  ( a -f-  6 -f-  c) . 
fjc=altitude  on  side  c. 
»»c=median  on  side  c. 
Jc=biseetor  of  angle  opposite 
side  c. 

landm  = line  segments. 
P=perimeter. 

<S«=side  of  a regular  polygon 

of  n sides.  &i 


P = radius  of  a circle. 

D= diameter  of  a circle. 

C= circumference  of  acirele. 
r— radius  of  an  inscribed 
circle. 

—^r  approx,  (or  3. 1416 — ). 
ir=area. 

6= base  of  a triangle. 
li  = altitude  of  a triangle, 
and  &2=bases  of  a trapezoid. 


LENGTHS  OF  LINES 


1.  In  a right  triangle,  C being  the  right  angle, 
c2=a2  + 62. 


Art.  346. 


Art.  342. 


2.  In  a right  triangle,  l and  m being  the  projections  of  a and  b 
on  c,  and  h,  the  altitude  on  c, 

h2  = lXm. 

a?  = lX.c,  V2  = m'Xc. 

3.  In  an  oblique  triangle,  m being  the  projection  of  5 on  c, 

if  a is  opposite  an  acute  Z,  a'i  = 62  + c2  — 2 cX»*.  Art.  349. 
if  a is  opposite  an  obtuse  Z , a2  = &2  — c2  — (—  2 c X m-  Art.  350. 


4.  hc=ti/ s (s — a)  (s  — &)  (s — c)* 


5.  mc=-k\/2  (a2  + &2) — c2. 

2 

6.  tc  = — — — c). 

CL  “r  0 


Art.  393. 
Art.  353. 
Art.  363. 


7.  If  l and  m are  the  segments  of  c made  by  the  bisector  of  the 
angle  opposite,  a ;b=l  :m.  Arts.  332,  336. 

(538) 


FORMULAS  OF  PLANE  GEOMETRY 


539 


8.  If  l and  m are  the  segments  of  a line,  a,  divided  in  extreme  and 


mean  ratio,  and  l > m,  a . i~i  : nli 

„ _ . , < P : P'=a  : a'. 

9.  In  similar  polygons,  < , , ,, 

U:  a'  = b : V . 

10.  In  circles,  C : C'  = B : B' ; also  C : C'  = D : IP. 

11.  C=2  ttH,  or  C=  wD. 

. central  angle.  . „ 

12.  An  arc  = i80b~~~X  B' 

13.  In  inscribed  regular  polygons, 

S2n= V B (2  B — j/4  B‘‘ — Sn). 


Art.  370. 

Art.  341. 
Art.  321. 

Art.  442. 

Art.  444. 

Art.  445. 

Art.  467. 


AREAS  OF  PLANE  FIGURES 
1.  In  a triangle,  K=i  bh. 


2.  In  a triangle,  K=\/s  (s — a)  (s  — b)  ( s — c). 


3.  In  an  equilateral  triangle,  K= 


oV  3 


4.  In  a parallelogram,  K—bX.  h. 

5.  In  a trapezoid,  K=i  h ( bl  4-  b2). 

6.  In  a regular  polygon,  rXE 

7.  In  a circle,  K=^B2  or  K=  J 

8.  In  a sector  of  a circle,  K=i  B X arc, 

central  Z 


or  K=- 


360° 


X lrB‘l 


Ex. 


9.  In  a segment  of  a circle,  lT=seetor  ± A formed  by 
and  radii  of  the  segment. 

10.  In  a circular  ring,  K=t*  ( B 2 — Bn). 

11.  In  any  two  similar  plane  figures, 

K : A'=a2  : a'2 ; 
also  a : a'=\/K  : |X K' . 

12.  In  two  circles,  K : E'=B2  : Bn  = D2  : Dn=C2  : Cn; 

also  B : B'  = D : D'=C  : C'  = \/K  : j/2T. 


Art.  389. 
Art.  393. 
4,  p.  257. 
Art.  385. 
Art.  394. 
Art.  446. 
Art.  449. 
Art.  453. 
Art.  453. 
the  chord 

Art.  449. 

Art.  399. 
Art.  314. 
Art.  452. 

Art.  314. 


FORMULAS  OF  SOLID  GEOMETRY 


SYMBOLS 


B,  &= areas  of  the  lower  and  up- 
per bases  of  a frustum. 
P=lateral  edge  (or  element) ; 
or 

= spherical  excess. 

JT=  altitude. 

I,  b,  7i.=length,  breadth,  height. 
Z=slant  height. 

M=are&  of  midseetion. 


P= perimeter  of  right  section. 
P,  p = perimeters  of  lower  and  up- 
per bases  of  a frustum 
r,  r'= radii  of  bases. 

5=  area  of  lateral  surface;  or 
= area  of  surface  of  sphere, 
etc. 

P=area  of  total  surface. 
F=volume. 


FORMULAS  FOR  AREAS 

1.  In  a prism,  S=EyP. 

2.  In  a regular  pyramid,  5=i  LyP. 

3.  In  a frustum  of  a regular  pyramid,  S=i  ( P-\-p ) L. 

4.  In  a cylinder  of  revolution,  £=2  ttRII. 

T=  2 rr R (R  + H). 

5.  In  a cone  of  revolution,  S=^RL. 

T=-*R  ( L + R ). 

6.  In  a frustum  of  a cone  of  revolution,  S=^L  {B  + r). 
-7.  In  a sphere,  5=4  ^Rr,  or 

8.  In  a zone,  5=2  ^RH. 

ttR-A 


9.  In  a lune,  5=- 


90 


10.  In  a spherical  triangle,  5= 

11.  In  a spherical  polygon,  5= 


■n-R-E 
180  ' 
TrR?E 
180 


Art.  608. 
Art.  641. 
Art.  643. 

Art.  697. 
Art.  697. 
Art.  721. 
Art.  721. 
Art.  727. 
Art.  810. 
Art.  S13. 

Art.  817. 
Art,  822. 

Art.  824. 


FOKMULAS  OF  SOLID  GEOMETRY 


541 


FORMULAS  FOR  VOLUMES 

1.  In  a prism,  V=BX  S.  Art.  628. 

2.  In  a parallelopiped,  F=ZX&Xb.  Art.  626. 

3.  In  a pyramid,  V=§  BX3-  Art.  651. 

4.  In  a frustum  of  a pyramid,  V=i  H (B  -\-  6 + l/ Bb).  Art.  656. 

5.  In  a prismatoid,  F=i  H (R  + &+4AI).  Art.  663. 

6.  In  a cylinder,  V=BXS-  Art.  698 • 

7.  In  a cylinder  of  revolution,  V=x’B2H.  Art.  699. 

8.  In  a cone,  V=iBXS.  Art.  722. 

9.  In  a circular  cone,  V—i  Art.  723. 

10.  In  a frustum  of  a cone,  V=§  H {B  + 5 + l/ Bb).  Art.  729. 

11.  In  a frustum  of  a cone  of  revolution, 

V=i  ttH  (R2  + r-  + Er).  Art.  730. 

12.  In  a sphere,  F=f  xB3,  or  F=|  xjf.  Art.  832. 

13.  In  a spherical  sector,  F=f  xR2H.  Art.  836. 

14.  In  a spherical  segment  of  two  bases, 

V=i  {xr2  + t rr/2)  H + i xH3.  Art.  837. 


15.  In  a spherical  segment  of  one  base,  V=xR2  (R  — iR). 

Art.  838. 


CONSTANTS 


1 acre =43,560  sq.  ft. 

1 bushel=2150.42  cu.  in. 
1 gallon  = 231  cu.  in. 

>/2  = 1.4142  + 

l/3  = 1.7321  — 

GO 


1^2  = 1.2599  + 

^3  = 1.4422  + 
i=.3183  + 

7 r 

l/  7T  = 1.7725  — 

-j-=0.5642  + 
yx 


542 


GEOMETRY.  APPENDIX 


SUMMARY  OF  METRIC  SYSTEM 


TABLE  FOR  LENGTH 


10  millimeters  (mm.) 
10  cm. 

10  dm. 

10  m. 

10  Dm. 

10  Hm. 

10  Km. 


= 1 centimeter  (cm.) 

= 1 decimeter  (dm.) 

= 1 meter  (m.). 

= 1 Dekameter  (Dm.) 

= 1 Hektometer  (Hm.) 
= 1 Kilometer  (Km.) 

= 1 Myriameter  (Mm.) 


Similar  tables  are  used  for  the  unit  of  weight,  the  gram ; for  the 
unit  of  capacity,  the  liter ; for  the  unit  of  land  measure,  the  are;  and 
for  the  unit  of  wood  measure,  the  stere. 


TABLE  FOR  SQUARE  MEASURE 

100  sq.  mm.  = l sq.  cm. 

100  sq.  cm.  =1  sq.  dm.,  etc. 

TABLE  FOR  CUBIC  MEASURE 

1000  cu.  mm.  = l cu.  cm. 

1000  cu.  cm.  =1  cu.  dm.,  etc. 

A liter  =1  cu.  dm. 

A p?-a>«  = weight  of  1 cu.  cm.  of  water  at  39.2°  Fahrenheit^ 
An  are  =100  sq.  m. 

A stere  =1  cu.  m. 

EQUIVALENTS 

1 meter  =39.37  inches. 

1 liter  =1.057  liquid  quarts, 

or  .9581  dry  quarts. 

1 kilogram  = 2. 2046  lbs.  av. 

1 hektare  =2.471  acres. 

1sq.m.  ="1550—  sq.  in. 


INDEX  OE  DEFINITIONS  AND  FORMULAS 


PAGE 

PAGE 

Abbreviations  . . . .8, 

231 

Angle  of  lune 

. 452 

Algebraic  analysis 

229 

polyhedral  . 

. 349 

Algebraic  me.thod 

91 

re-entrant 

. 74 

Alternation 

180 

reflex  .... 

. 16 

Altitude  of  cone  .... 

412 

right  .... 

. 15 

of  cylinder 

403 

salient  .... 

. 73 

of  frustum  of  cone 

414 

sides  of  . 

. 14 

of  frustum  of  pyramid 

378 

spherical 

. 433 

of  parallelogram  . 

67 

straight 

. 15 

of  prism 

361 

tetrahedral  . 

. 349 

of  pyramid  . . . . 

377 

trihedral 

. 349 

of  spherical  segment  . 

462 

vertex  .... 

. 33 

of  trapezoid  . . . . 

67 

vertex  of 

. 14 

of  triangle 

33 

Angles,  adjacent  . 

. 16 

of  zone  

452 

alternate-interior  . 

55 

Analysis,  solution  by 

97 

complementary . 

. 16 

Angl  e 

14 

exterior  .... 

. 55 

acute  

15 

exterior-interior 

55 

at  center  of  regular  poly- 

homologous . 

. 34 

g°n 

264 

interior  .... 

. 55 

central  

105 

of  polygon  . 

. 73 

dihedral  

337 

of  quadrilateral 

. 66 

exterior  in  triangle  . 

32 

of  spherical  polygon 

. 438 

formed  by  a rotating 

of  triangle  . 

. 32 

straight  line 

17 

opposite-interior 

. 32 

formed  by  two  curves 

433 

supplementary  . 

. 16 

inscribed  in  circle 

105 

vertical  .... 

. 16 

inscribed  in  segment  of 

Antecedents 

. 177 

circle  

105 

Apothem  .... 

. 263 

oblique 

16 

Appolonius 

. 493 

obtuse 

15 

Arabs  

543 


544 


INDEX  OF  DEFINITIONS  AND  FORMULAS 


PAGE 

PAGE 

Arc 

. 103 

Center  of  symmetry 

. 293 

formula  for 

. 275 

Centroid 

. 83 

major 

. 104 

Chord  .... 

. 103 

minor 

. 104 

Circle  .... 

103,  496 

Archimedes 

497,498 

arc  of 

. 103 

Arcs,  conjugate  . 

. 104 

center  of 

. 103 

Area  of  surface  . 

. 231 

circumference  of 

. 103 

Aryabhatta  .... 

. 497 

circumference  of, 

formula 

August,  E.  F. 

. 498 

for 

. 274 

Auxiliary  lines 

93,  170 

circumscribed  . 

. 105 

Axiom 

. 19 

diameter  of 

. 103 

Axioms,  general  . 

19,  20 

formula  for  area 

of 

. 276 

geometric  .... 

20,  21 

great 

. 427 

Axis  of  circle  of  sphere  . 

. 427 

inscribed 

. 105 

of  circular  cone 

. 413 

radius  of 

. 103 

of  regular  pyramid 

. 378 

sector  of 

. 104 

of  sphere  .... 

. 427 

small 

. 427 

of  symmetry 

. 293 

Circles,  concentric 

. 105 

escribed 

. 162 

Babylonians  .... 

. 497 

tangent  . 

. 105 

Base  of  cone  .... 

. 412 

Circum-eenter  of  triangle 

. 81 

of  isosceles  triangle  . 

. 33 

Circumference 

14,  103 

of  pyramid 

. 377 

Classification  of  polyhed 

ons  360 

of  spherical  pyramid 

. 461 

of  triangles 

32,  33 

of  spherical  sector 

. 461 

Commensurable 

. 125 

of  triangle  .... 

. 33 

Complement 

. 16 

Bases  of  cylinder 

. 403 

Composition 

. 1S1 

of  frustum  of  cone 

. 414 

and  division 

. 182 

of  frustum  of  pyramid 

. 378 

Conclusion 

. 24 

of  parallelogram  . 

. 67 

Concurrent  lines  . 

. 80 

of  prism  ..... 

. 361 

Coneylic  points 

. 105 

of  spherical  segment  . 

. 462 

Cone  .... 

. 412 

of  trapezoid 

. 67 

altitude  of 

. 412 

of  zone  

. 452 

axis  of  . 

. 413 

Bisector  of  triangle  . 

. 34 

base  of  . 

. 412 

Bodies,  the  Three  Bound 

. 498 

circular  . 

. 413 

circular,  formula 

for  vol- 

Center  of  circle  . 

. 103 

ume  of 

. 41S 

of  regular  polygon 

. 263 

circular,  formulas  for 

at- 

of  sphere  .... 

. 425 

eral  and  total  area 

. 417 

INDEX  OF  DEFINITIONS  AND  FORMULAS 


545 


PAGE 

Cone,  elements  of 

. 412 

lateral  surface  of  . 

. 412 

oblique  circular 

. 413 

of  revolution 

. 413 

right  circular  . 

. 413 

vertex  of  ... 

. 412 

■ Cones,  similar 

. 413 

Congruent  figures 

. 13 

Conical  surface  . 

. 412 

directrix  of 

. 412 

element  of  ... 

. 412 

generatrix  of 

. 412 

nappes  of  ... 

. 412 

vertex  of  ... 

. 412 

Conjugate  arcs 

. 104 

•Consequents  .... 

. 177 

Constant 

. 126 

1 

geometrical  .... 

. 143 

Constants 

. 541 

Continued  proportion 

. 177 

1 

Continuity,  principle  of  . 

. 485 

Converse  of  a theorem  . 

. 24 

Corollary 

. 23 

Dube 

. 363 

I Curved  spaces 

. 488 

i | Curved  surface  . . 

. 17 

1 1 Cylinder 

. 403 

1 altitude  of  ... 

. 403 

' 

bases  of  .... 

. 403 

circular  

. 404 

2 

circular  properties  of  . 

. 405 

11) 

i elements  of  . 

. 403 

13' 

1 lateral  surface  of  . 

. 403 

,12 

oblique 

. 404 

13 

1 of  revolution  „ 

. 404 

| of  revolution,  formulas  for 

118 

1 lateral  and  total  areas 

of 

. 409 

411 

1 right 

. 404 

| right  circular  . 

. 404 

PAGE 

Cylinder,  right  section  of  , 405 

section  of 405 

Cylinders  of  revolution,  sim- 
ilar   404 

Cylindrical  surface  . • . 403 

directrix  of 403 

element  of 403 

generatrix  ......  403 

Decagon 74 

Degree  of  arc 131 

angular 17 

spherical 452 

Demonstration,  indirect  . 25, 96 
Determined  plane  . 319, 320 

Diagonal  of  polygon  . . .73 

of  polyhedron  ....  360 
of  quadrilateral  ...  66 

Diameter  of  circle  . . . 103 

of  sphere 425 

Dihedral  angle  ....  337 

edge  of 337 

faces  of 337 

plane  angle  of  ...  338 

right 337 

Dihedral  angles,  adjacent  . 337 

equal 337 

vertical 337 

Dimensions 12 

Directrix  of  conical  surface.  412 
of  cylindrical  surface  . 403 , 

Discussion 147 

Distance  from  point  to  line.  35 
Distance  from  point  to  plane  328 
on  surface  of  sphere  . . 428 

polar,  of  great  circle  . . 429 

polar,  of  small  circle  . . 429 

Division 181 

Dodecagon 74 

Dodecahedron  ....  360 


546  INDEX  OF  DEFINITIONS  AND  FORMULAS 


Duality,  Principle  of 

PAGE 

. 486 

Edge  of  dihedral  angle  . 

. 337' 

Edges  of  polyhedral  angle 

. 349 

of  polyhedron  . 

. 360 

Egyptians 


. . 490,  491,  492,  494,  497 

Elements  of  conical  surface  412 


of  cylindrical  surface  . 

. 403 

Enunciation,  general 

. 25 

particular  .... 

. 25 

Equal  figures  .... 

. 13 

Equivalent  figures 

13,  231 

solids 

. 363 

Euclid 

. 491 

Eudoxus  . . . 493,  496,  498 

Euler 

. 497 

European  

. 490 

Extreme  and  mean  ratio 

. 217 

Extremes 

. 177 

Faces  of  dihedral  angle  . 

. 337 

of  polyhedral  angle 

. 349 

of  polyhedron  . 

. 360 

Figure,  curvilinear  . 

. 13 

geometrical  .... 

. 13 

mixtilinear  .... 

. 13 

plane 

. 17 

rectilinear  .... 

. 13 

Figures,  rectilinear  . 

. 494 

Formulas  of  Plane  Geometry, 


for  lengths  of  lines.  538,  539 
of  Plane  Geometry,  for 
areas  of  plane  figures  . 539 
of  Solid  Geometry,  for 

areas 540 

of  Solid  Geometry,  for 

volumes 541 

Foot  of  perpendicular  . . 15 

of  line 320 


Fourth  proportional 

PAGE 

. . 177 

Frustum  of  cone 

. . 414 

altitude  of  . 

. . 414 

bases  of 

. . 414 

formula  for  lateral 

area 

of  .... 

. . 420 

formula  for  volume 

of  . 422 

lateral  surface  of  . 

. . 414 

slant  height  of 

. . 414 

Frustum  of  pyramid 

. . 378 

altitude  of  . 

. . 378 

bases  of 

. . 378 

slant  height  of 

. . 378 

Generatrix  of  conical 

sur- 

face  .... 

. . 412 

of  cylindrical  surface 

. 403 

Geometric  solid  . 

. . 11 

Geometry  .... 

. . 11 

epochs  in  development  of  490 

history  of  . 

.490-498 

modern  .... 

. . 485 

Non-Euclidean  . 

488,  498 

origin  of  . 

. . 490 

plane  .... 

. . 18 

projective 

. . 485 

solid 

. 18,  319 

Gerard  .... 

. . 49S 

Greeks  . 490,  492, 

493,  494 

Harmonic  division 

. . 193 

Heptagon  .... 

. . 74 

Hero 

. . 496 

Hexagon  .... 

. . 74 

Hexahedron 

. . 360 

Hindoos  . . . 490, 

492,  497 

Hippasius  .... 

. . 498 

Hippocratus  . . 491, 

492,  496 

History  of  geometry  . 

490-493 

Homologous  angles  . 

, . 34 

sides  .... 

. . 34 

INDEX  OF  DEFINITIONS  AND  FORMULAS 


547 


PAGE 

PAGE 

Homology,  Principle  of 

. 487 

Line  divided  internally  . 

• 

198 

Hyperspace  .... 

. 488 

of  centres  .... 

. 

120 

Hypotenuse  .... 

. 33 

parallel  to  plane  . 

321 

Hypothesis  .... 

. 24 

perpendicular  to  plane 

320 

straight  .... 

. 

13 

Icosahedron  .... 

. 360 

In-center  of  triangle 

. 80 

Lines,  auxiliary 

93,  170 

Inclination  of  line  to  plane  347 

concurrent  .... 

80 

Incommensurable 

. 125 

Lobatchewsky 

498 

ratio 

. 125 

Locus  

52 

Inference,  immediate 

. 23 

Logical  methods  . 

492 

Inversion 

. 180 

Lune  .... 

452 

Isoperimetric  figures 

. 286 

angle  of  .... 

45? 

formula  for  area  of, 

in 

J ews 

. 497 

spherical  degrees 

457 

formula  for  area  of. 

in 

Lateral  area  of  frustum 

of 

square  units  of  area 

. 

457 

pyramid,  formula  for 

. 379 

of  prism  .... 

. 361 

Major  arc  of  circle  . 

104 

of  pyramid  .... 

. 377 

Magnitudes,  commensurable 

125 

Lateral  edges  of  prism  . 

. 361 

incommensurable  . 

125 

of  pyramid 

. 377 

Maximum  ....  141, 

285 

Lateral  faces  of  prism  . 

. 361 

Mean  proportional 

177 

of  pyramid 

. 377 

Means 

177 

Lateral  surface  of  cone 

. 412 

Mechanical  methods 

493 

of  cylinder 

. 403 

Median  of  trapezoid  . 

67 

of  frustum  of  cone 

. 414 

of  triangle 

34 

Legs  of  isosceles  triangle 

..  33 

Menelaus  . . . . 

498 

of  right  triangle  . 

. 33 

Method,  algebraic 

91 

of  trapezoid 

. 67 

of  limits  .... 

126 

Limit 

. 126 

Methods,  logical  . 

492 

Limits,  method  of 

. 126 

mechanical  .... 

493 

Line 

. 12 

of  numerical  computa- 

broken  

. 13 

tions 

304-5 

curved  

. 13 

rhetorical  .... 

491 

divided  externally 

. 198 

Metric  system,  summary 

of 

542 

divided  harmonically 

. 198 

Minor  arc  of  circle  . 

104 

divided  in  extreme  and 

Minute 

17 

mean  ratio  . 

. 217 

Modern  geometry 

. 

485 

548  INDEX  OF  DEFINITIONS  AND  FORMULAS 


PACE 

PAGE 

Nappes  of  cone  . 

. . 412 

Pole 

427 

Negative  quantities  . 

485,  486 

Polygon  .... 

73 

N-gon 

. . 74 

angles  of 

73 

Non-Euclidean  geometry 

circumscribed  . 

105 

488,  498 

concave  .... 

73 

Numerical  measure  . 

. . 124 

convex  .... 

73 

computation,  methods  of 

diagonal  of 

73 



304-305 

equiangular 

73 

equilateral  . 

73 

Octagon  .... 

. . 74 

inscribed 

105 

Octahedron 

. . 360 

perimeter  of 

73 

Opposite  of  a theorem 

. . 24 

regular  .... 

261 

Origin  of  geometry  . 

. . 490 

regular,  angle  at  center 

of  264 

Ortho-center 

. . 82 

regular,  apothem  of 

263 

regular,  center  of  . 

263 

Parallel  lines 

. . 56 

regular,  radius  of 

263 

Parallelogram 

. . 66 

sides  of  ... 

73 

altitude  of  . . 

. . 67 

spherical 

438 

bases  of  ... 

. . 67 

vertices  of  . 

73 

Parallelopiped 

. . 362 

Polygons,  mutually 

equi- 

right  .... 

. . 362 

angular 

74 

rectangular 

. . 363 

mutually  equilateral 

74 

Parallel  planes  . 

. . 321 

similar  .... 

190 

Pentagon  .... 

. . 74 

symmetry  of 

293 

Pentedecagon  . 

. . 74 

Polyhedral  angle  . 

349 

Perimeter  of  polygon 

. . 73 

convex  .... 

349 

of  quadrilateral 

. . 66 

edges  of  ... 

349 

of  triangle  . 

. . 32 

face  angles  of  . 

349 

Perpendicular  lines  . 

. . 15 

faces  of  . 

349 

planes  .... 

. . 338 

vertex  of 

349 

Pi  (tt)  .... 

. . 274 

Polyhedral  angles,  equal 

350 

Plane  .... 

17,319 

symmeti'ical 

350 

figure  .... 

. . 17 

vertical  .... 

350 

determination  of  . 

319,  320 

Polyhedron 

360 

Planes,  parallel  . 

. . 321 

convex  .... 

360 

Plato  

491, 493 

diagonal  of  . 

360 

Point 

. . 12 

edges  of  . . . . 

360 

Points,  concylic  . 

. . 105 

faces  of  . 

360 

Polar  distance  of  circle 

. . 429 

regular  .... 

391 

triangle  .... 

. . 441 

section  of 

360 

INDEX  OF  DEFINITIONS  AND  FORMULAS 


549 


Polyhedron,  vertices 

PAGE 

of  . 360 

Polyhedrons 

. . 497 

classification  of 

. . 360 

similar  . 

. . 396 

Postulate  . 

22 

of  solid  geometry  . 

. . 320 

Postulates 

. 22,  146 

logical  . 

22 

Prism 

. . 361 

altitude  of  . 

. . 361 

bases  of  ... 

. . 361 

circumscribed  about  cyl- 

inder  .... 

. 405 

inscribed  in  cylinder  . . 405 

lateral  area  of  . 

. . 361 

lateral  edges  of 

. . 361 

lateral  faces  of 

. . 361 

oblique  .... 

. . 362 

quadrangular  . 

. . 362 

regular  .... 

. . 362 

right  .... 

. . 362 

right  section  of 

. . 361 

triangular  . 

. . 362 

truncated 

. . 362 

Prismatoid 

. . 339 

bases  of  .... 

. . 389 

formula  for  volume 

of  . 390 

Prismoid  .... 

. . 3S9 

Problem  .... 

. . 23 

Projection  of  line  on 

line  . 205 

of  line  on  plane  . 

. . 33S 

of  point  on  line 

. . 205 

of  point  on  plane  . 

. . 338 

Proof 

. . 23 

by  superposition  . 

. . 25 

forms  of  ... 

. . 25 

methods  of  . 

. . 25 

Proportion 

. . 177 

continued 

. . 177 

terms  of  ... 

..  . 177 

PAGE 


Proportional,  fourth  . 

. . 177 

mean  .... 

. . 177 

third  .... 

. . 17S 

Proposition 

. . 23 

Pyramid  .... 

. . 377 

altitude  of  . 

. . 377 

axis  of  ... 

. . 378 

base  of  ... 

. . 377 

circumscribed  - about 

cone  414 

frustum  of  . 

. . 378 

inscribed  in  cone  . 

. . 414 

lateral  area  of  . 

. . 377 

lateral  edges  of 

. . 377 

lateral  faces  of 

. . 377 

quadrangular  . 

. . 377 

regular  .... 

. . 377 

regular,  slant  height 

of  . 378 

spherical 

. . 461 

triangular  . 

. . 377 

truncated 

. . 378 

vertex  of 

. . 377 

Pythagoras  . 491,  492, 

495,  497 

Quadrant  of  circle 

. . 103 

Quadrilateral  . 

. . 66 

angles  of 

. . 66 

perimeter  of 

. . 66 

sides  of  . . .'  . 

. . 66 

vertices  of  . 

. . 66 

Radius  of  circle  . 

. . 103 

of  regular  polygon 

. . 263 

of  sphere 

. . 425 

Ratio 

. . 125 

of  similitude 

. . 190 

Reciprocally  proportional  . 211 
Reciprocity,  principle  of  . 486 

Rectangle  .... 

. . 60 

Rectilinear  figures  . 

. . 494 

Rhetorical  methods  . 

. . 491 

Rhomboid 

66 

550 


INDEX  OF  DEFINITIONS  AND  FORMULAS 


Rhombus 

PAGE 
. 66 

Right  section  of  cylinder 

. 405 

of  prism  .... 

. 361 

Romans  ....  494, 496 

Round  Bodies,  The  Three 

. 498 

Scholium 

. 23 

Secant  line  .... 

. 104 

Second 

. 17 

Sect 

. 14 

Section  of  polyhedron  . 

. 360 

Sector  of  circle 

. 104 

of  circle,  formula  for 

. 277 

of  sphere  .... 

. 461 

Sectors,  similar  . 

. 275 

Segment  of  circle 

. 194 

of  line 

. 14 

of  sphere  .... 

. 462 

Segments,  similar 

. 275 

Semicircle 

. 104 

.Semicircumference 

. 103 

Sides  of  angle 

. 14 

of  polygon  .... 

. 73 

of  quadrilateral 

. 66 

of  triangle  .... 

. 32 

Similar  cones  of  revolution  413 

cylinders  of  revolution 

. 404 

figures 

. 63 

polygons  .... 

. 190 

polyhedrons 

. 396 

sectors  of  circles  . 

. 275 

segments  of  circles 

. 275 

Slant  height  of  cone  . 

. 413 

of  frustum  of  cone  . 

. 414 

of  frustum  of  pyramid 

. 378 

of  \ogular  pyramid 

. 377 

Solid 

11, 12 

geometry  .... 

18,319 

Solids,  equivalent 

. 363 

geometric  . 

. 11 

Solids,  physical  . 

PAGE 
. . 11 

Spaces,  curved 

. . 488 

Sphere 

. . 425 

axis  of  ... 

. . 427 

center  of 

. . 425 

circumscribed  about  poly- 

hedron 

. . 432 

diameter  of 

. . 425 

formula  for  area  of 

sur- 

face  of  . 

. . 455 

formula  for  volume 

of  . 463 

great  circle  of  . 

. . 427 

dro 


a . 432 
. 427 
. 425 
. 427 
. 433 
. 452 
444,  4C0 
. 438 


inscribed  in  polyli 
poles  of  . 
radius  of 
small  circle  of  . 

Spherical  angle  . 
degrees  . 
excess 

Spherical  polygon 

angles  of 438 

sides  of 438 

vertices  of 438 

Spherical  pyramid  . . .461 

base  of 461 

vertex  of  . . . .461 

Spherical  sector  . . . .461 

base  of 461 

formula  for  volume  of  . 464 

Spherical  segment  . . . 462 

altitude  of 462 

bases  of 462 

formulas  for  volume  of  . 465 

of  one  base 462 

Spherical  triangle  . . . 438 

bi-rectangular  . . . 443 

formula  for  area  of,  in 

square  units  of  area  . 460 

tri-rectansrular  . . . 443 


INDEX  OF  DEFINITIONS  AND  FORMULAS  551 


PAGE 

PAGE 

Spherical  triangles,  supple- 

Trapezoid, isosceles  . 

. 67 

mental  

442 

median  of  ... 

. 67 

symmetrical  . 

444 

Triangle 

32,  74 

Spherical  wedge  . 

461 

acute  

. 33 

Spherid  ...... 

452 

altitudes  of 

. 33 

Square  

66 

base  of  

. 33 

Straight  line 

13 

bisectors  of  . 

. 34 

Superposition  . . . . 

25 

bisectors  of,  formula 

for  213 

Supplement 

16 

classifications  of  . 

32,  33 

Surface 

12 

equiangular 

. 33 

area  of  

231 

equilateral  .... 

. 32 

conical 

413 

isosceles  .... 

. 32 

curved 

17 

medians  of  . 

. 34 

cylindrical 

403 

obtuse  

. 33 

unit  of 

234 

polar 

. 441 

Symbols 

8 

right 

. 33 

Symmetry,  axis  of  . 293,  294 

scalene  

. 32 

center  of  293,  294 

spherical  .... 

. 438 

center  of,  for  a polygon  . 

293 

vertex  angle  of 

. 33 

vertex  of  ... 

. 33 

Tangent  circles  .... 

105 

Trihedral  angle  . 

. 349 

Tangent,  common  external  . 

120 

bi-rectangular  . 

. 350 

common  internal  . 

120 

isosceles  .... 

. 350 

line  to  sphere  .... 

425 

rectangular 

. 350 

plane  to  cone  .... 

413 

tri-rectangular 

. 350 

plane  to  cylinder  . 

404 

plane  to  sphere 

426 

Ungula 

. 461 

spheres 

426 

Unit  of  measure  . 

. 124 

Terms  of  a proportion  . 

177 

of  surface  . 

. 231 

Tetrahedral  angle 

349 

of  volume  .... 

. 363 

Tetrahedron 

360 

Units  of  angle 

. 17 

Thales  . . . 491,495, 

496 

of  spherical  surface  . 

. 452 

Theorem 

23 

Theory  of  limits  .... 

126 

Variable 

. 126 

Third  proportional  . 

178 

Vertex  angle  ... 

. 33 

Transversal  

55 

Vertex  of  angle  . 

. 14 

Trapezium 

66 

of  cone  

. 412 

Trapezoid 

66 

of  polyhedral  angle  . 

. 349 

altitude  of  . . ... 

67 

of  pyramid  .... 

. 377 

bases  of  . w 

67 

of  spherical  pyramid  . 

. 461 

552  INDEX  OF  DEFINITIONS  AND  FORMULAS 


Vertex  of  triangle  . 

PAGE 

. 33 

Vertices  of  polygon  . 

. 73 

of  polyhedron  . 

. 360 

of  quadrilateral 

. 66 

of  spherical  polygon  . 

. 438 

of  triangle  .... 

. 32 

Volume  of  solid  . 

. 363 

Wedge,  spherical  . 

PAGE 

. . 461 

Xenodorus 

. . 496 

Zone 

. . 452 

altitude  of  . 

. . 452 

bases  of  . . . 

. 452 

of  one  base  . 

. . 452 

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